ORTON'S 
1  LIGHTNING  CALCULATOR, 

ACGOMTANT'S  ASSISTANT, 


iViLiiJl«iiiiiiiiiiiiiii«iiiil 

of  u1  1  within 

bne  having  the 
•  Fig1" 


BY  HOYm  ORTON, 

705  JAYNE  STREET,  PHILADELPHIA.. 


ENERGY  IS  THE  PRICE  Or  ^ 


IT,  B,— Any  infringement  upon  the  copyright  of  this  book  will  be 
prosecuted  to  the  fullest  extent  of  the  law. 


Address  all  Orders  for  this  Book  to 

T.  K.  COLLINS,  705  Jayne  St.,  Philadelphia. 

FOJR  SALE  HY ALL  WHOLESALE  BOOKSELLERS. 
Single  Copies,  by  Mail,  One  Dollar, 


IN  MEMORIAM 
FLOR1AN  CAJORI 


LIGHTNING  CALCULATOR, 


ACCOUNTANT'S  ASSISTANT. 


THE  SHORTEST,  SIMPLEST,  AND  MOST  RAPID  METHOD  OP  COMPUTING 

NUMBERS,    ADAPTED  TO    EVERT    KIND   OF   BUSINESS,    AND 

WITHIN  THE  COMPREHENSION  OF  EVERY  ONE  HAVING 

THE  SLIGHTEST  KNOWLEDGE  OF  FIGURES. 


BY  HOY  D.  ORTOX, 

705  JAYNE    STREET,  PHILADELPHIA. 


ENERGY  IS  THE  PRICE  OF  SUCCESS. 


N.  B.— Any  infringement  upon  the  copyright  of  this  book  will  be 
prosecuted  to  the  fullest  extent  of  the  law. 


Address  all  Orders  for  this  book  to 

T,  K,  COLLINS,  705  Jayne  Street,  Philadelphia, 
For  Sale  by  all  Wholesale  Booksellers, 

Single  Copies,  by  Mail,  One  Dollar. 
1869. 


Entered  according  to  Act  of  Congress,  in  the  year  1866,  by 
HOY  D.  ORTON, 

fcl  the  Clerk's  Office  of  the  District  Court  of  the  United  State* 
for  the  Southern  District  of  Ohir. 


COiLINS,  PRINTER. 


INTRODUCTION. 

QUANTITY  is  that  which  can  be  increased  or 
diminished  by  augments  or  abatements  of  homo* 
geneous  parts.  Quantities  are  of  two  essential 
kinds,  Geometrical  and  Physical. 

1.  Geometrical  quantities  are  those  which  occupy 
space ;  as  lines,  surfaces,  solids,  liquids,  gases,  etc. 

2.  Physical  quantities  are  those  which  exist  in 
the  time,  but  occupy  no  space ;  they  are  known  by 
their  character  and  action  upon  geometrical  quan- 
tities, as  attraction,  light,  heat,  electricity  and  mag- 
netism,  colors,  force,  power,  etc. 

To  obtain  the  magnitude  of  a  quantity  we  com- 
pare it  with  a  part  of  the  same ;  this  part  is  im- 
printed in  our  mind  as  a  unit,  by  which  the  whole 
is  measured  and  conceived.  No  quantity  can  be 
measured  by  a  quantity  of  another  kind,  but  any 
quantity  can  be  compared  with  any  other  quantity, 
and  by  such  comparison  arises  what  we  call  caZcu- 
lation  or  Mathematics 


IV  INTRODUCTION. 

MATHEMATICS. 

MATHEMATICS  is  a  science  by  which  the  com- 
parative value  of  quantities  are  investigated ;  it  ia 
divided  into  : 

1.  ARITHMETIC,  that  branch  of    Mathematics 
which  treats  of  the  nature  and  property  of  num- 
bers ;  it  is  subdivided  into  Addition,  Subtraction, 
Multiplication,  Division,  Involution,  Evolution  and 
Logarithms. 

2.  ALGEBRA,  that  branch  of  Mathematics  which 
employs  letters  to  represent  quantities,  and  by  that 
means    performs    solutions  without    knowing    or 
noticing  the  value  of  the  quantities.     The  subdi- 
visions of  Algebra  are  the  same  as  in  Arithmetic. 

3.  GEOMETRY,  that  branch  of  Mathematics  which 
investigates  the  relative  property  of  quantities  that 
occupies  space ;    its  subdivisions  are   Longemetry^ 
Planemetry,   Stereometry,    Trigonometry  and  Conic 
Sections. 

4.  DIFFERENTIAL-CALCULS,  that  branch  of  Math- 
ematics which  ascertains  the  mean  effect  produced 
by  group  of  continued  variable  causes. 

5.  INTEGRAL-CALCULS,  the  contrary  of  Differen- 
tial, or  that  branch  of  Mathematics  which  investi- 
gates the  nature  of  a  continued  variable  cause  that 
has  produced  a  known  effect. 


PREFACE. 


xj     . 

MATEJMATICAL   LAWS  are  the   acknowledged 
of  all  science.     Ever   since   the   streets  of 


Athens  i  ^sounded  with  that  historical  cry  of  "Eu- 
reka," emanating  from  one  of  antiquity's  greatest 
mathematicians,  the  science  has  been  steadily  pro- 
gressing. 

It  is  not  our  purpose,  in  this  small  work,  to  in- 
troduce any  of  the  higher  branches  of  mathematics, 
viz.:  Algebra,  Conic  Sections,  Calculus,  etc.  Our 
object  is  merely  to  present  to  the  public  a  system 
of  calculation  that  is  practical  to  every  business 
man.  It  consists  of  the  addition  of  numbers  on  a 
principle  entirely  different  from  the  one  ordinarily 
used.  In  the  practical  application  of  this  new  prin- 
ciple of  addition,  scarcely  any  mental  labor  is  re- 
quired, compared  with  the  principle  of  addition  set 
forth  in  standard  works.  The  superiority  we  claim 
for  this  principle  above  all  others,  is  this,  that  it 
requires  no  great  mental  exertion,  affording  the 


VI  PREFACE. 

greatest  facilities  to  the  calculator  in  the  additioi 
of  D umbers,  enabling  him  to  add  a  whole  day  with- 
out any  mental  fatigue ;  whereas,  by  the  ordinar} 
way,  it  is  very  laborious  and  fatiguing. 

Our  system  of  calculation  also  embraces  a  concise 
rapid,  and  at  the  same  time  practical  method  of 
Multiplication,  by  which  one  is  enabled  to  arrive 
at  the  product  of  any  number  of  figures,  multiplied 
by  any  number,  immediately,  without  the  use  of 
partial  products. 

This  small  work  also  embraces  the  shortest  and 
most  concise  method  for  the  computation  of  Interest 
ever  introduced  to  the  public.  Our  system  for  com- 
puting interest  is  entirely  different  from  any  rule 
ever  introduced,  for  the  computation  of  either  Sim- 
ple or  Compound  Interest.  A  student  having  gone 
no  further  than  Long  Division  in  Arithmetic,  can, 
by  our  rule,  calculate  Simple  or  Compound  Interest 
at  any  given  rate  per  cent.,  for  any  given  timo,  in 
one-tenth  of  the  time  that  the  best  calculators  will 
compute  it  by  the  rules  laid  down  in  other  books. 
By  using  our  rules,  you  can  entirely  avoid  the  use 
of  fractions,  and  save  the  calculation  of  75  to  100 
figures,  where  years,  months  and  days  are  giv^Q  OD 
a  note. 


ADDITION. 

To  BE  able  to  add  two,  three  or  four  columns  of 
figures  at  once,  is  deemed  by  many  to  be  a  Her- 
culean task,  and  only  to  be  accomplished  by  the 
gifted  few,  or,  in  other  words,  by  mathematical 
prodigies.  If  we  can  succeed  in  dispelling  this 
illusion,  it  will  more  than  repay  us ;  and  we  feel 
very  confident  that  we  can,  if  the  student  will  lay 
aside  all  prejudice,  bearing  steadily  in  mind  that  to 
become  proficient  in  any  new  branch  or  principle  a 
little  wholesome  application  is  necessary.  On  the 
contrary,  we  can  not  teach  a  student  who  takes  no 
interest  in  the  matter,  one  who  will  always  be  a 
drone  in  society.  Such  men  have  no  need  of  this 
principle. 

If  two,  three,  or  more,  columns  can  be  carried 
up  at  a  time,  there  must  be  some  law  or  rule  by 
which  it  is  done.  We  have  two  principles  of  Addi- 
tion; one  for  adding  short  columns,  and  one  for 
adding  very  long  columns.  They  are  much  alike, 
differing  only  in  detail.  When  one  is  thoroughly 
learned,  it  is  very  easy  to  learn  the  second.  By  a 
little  attention  to  the  following  example,  much  time 
in  future  will  be  saved. 


10         OKTON'S  LIGHTNING  CALCULATOR. 

ADDITION  OF  SHORT  COLUMNS  OF  FIGURES. 

ADDITION  is  the  basis  of  all  numerical  opera- 
tions, and  is  used  in  all  departments  of  business, 
To  aid  the  business  man  in  acquiring  facility  and 
accuracy  in  adding  short  columns  of  figures,  the 
following  method  is  presented  as  the  best: 

PROCESS. — Commence  at  the  bottom  of 

274 

g^g     the  right-hand  column,  add  thus:  16,  22, 
134     32 ;  then  carry  the  3   tens  to   the  second 
342     column ;  then  add  thus :  7,  14,  25 ;  carry 
the  2  hundreds  to  the  third  column,  and 
add   the   same  way:  12,  16,  21.     In  this 
2152     way  yOU  name  tbs  sum  of  two  figures  at 
once,  which  is  quite  as  easy  as  it  is  to  add  one 
figure  at  a  time.     Never  permit  yourself  for  once 
to  add  up  a  column  in  this  manner :  9  and  7  are 
16,  and  2  are  18   and  4  are  22,  and  6  are  28,  and 
4  are  32.     It  is  just  as  easy  to  name  the  result 
of  two  figures  at  once  and  four  times  as  rapid. 
The  following  method  is  recommended  for  the 

ADDITION  OF  LONG  COLUMNS  OF  FIGURES. 
In  the  addition  of  long  columns  of  figures 
which  frequently  occur  in  books  of  accounts,  in 
order  to  add  them  with  certainty,  and,  at  the 
same  time,  with  ease  and  expedition,  study  well 
the  following  method,  which  practice  will  render 
familiar,  easy,  rapid,  and  certain. 


ADDITION.  11 

THE    EASY    WAY    TO    ADD. 

EXAMPLE  2— EXPLANATION. 

Commence  at  9  to  add,  and  add  as  near  20  as  pos- 
sible, thus:  9+2+4+3=18,  place  the  8  to  tho 
right  of  the  3,  as  in  example;  commence  at  6  to  77 
add  6+4+8=18  ;  place  the  8  to  the  right  of  4 
the  8.  as  in  example ;  commence  at  6  to  add  6 
6+4+7=17  ;  place  the  7  to  the  right  of  the  36 
7,  as  in  example ;  commence  at  4  to  add  4+  9 
9+3=16  ;  place  the  6  to  the  right  of  the  3,  4 
as  in  example;  commence  at  6  to  add  6+4  7T 
+7=17;  place  the  7  to  the  right  of  the  7,  4 
as  in  example;  now,  having  arrived  at  the  6 
top  of  the  column,  we  add  the  figures  in  the  8* 
new  column,  thus:  7+6+7+8+8=36;  place  4 
the  right  hand  figure  of  36,  which  is  a  6,  6 
under  the  original  column,  as  in  example,  and  38 
add  the  left  hand  figure,  which  is  a  3,  to  the  4 
number  of  figures  in  the  new  column;  there  2 
are  5  figures  in  the  new  column,  therefore  9 
3+5=8;  prefix  the  8  with  the  6,  under  the  — 
original  column,  as  in  example  ;  this  makes  86 
86,  which  is  the  sum  of  the  column. 

Remark  1. — If,  upon  arriving  at  the  top  of  th< 
column,  there  should  be  one,  two  or  three  figures 
whose  sum  will  not  equal  10,  add  them  on  to  thfr 
Bum  of  the  figures  of  the  new  column,  never  placing 


12         ORTON'S  LIGHTNING  CALCULATOR. 

an  extra  figure  in  the  new  column,  unless  it  be  an 
excess  of  units  over  ten. 

Remark  2. — By  this  system  of  addition  you  can 
stop  any  place  in  the  column,  where  the  sum  of  the 
figures  will  equal  10  or  the  excess  of  10 ;  but  the 
addition  will  be  more  rapid  by  your  adding  as 
near  20  as  possible,  because  you  will  save  the  form- 
ing of  extra  figures  in  your  new  column. 
EXAMPLE— EXPLANATION. 

2+6+7=15,  drop  10,  place  the  5  to  the  right 
of  the  7;  6+5+4=15,  drop  10,  place  the  5  to 
the  right  of  the  4,  as  in  example;  8+3+7=18, 
drop  10.  place  the  8  to  the  right  of  the  7,  4 
as  in  example;  now  we  have  an  extra  figure,  78 
•which  is  4  j  add  this  4  to  the  top  figure  of  the  3 
new  column,  and  this  sum  on  the  balance  of  8 
the  figures  in  the  new  column,  thus:  4+8+  4* 
5+5=22;  place  the  right  hand  figure  of  22  5 
under  the  original  column,  as  in  example,  and  6 
add  the  left  hand  figure  of  22  to  the  num-  7* 
ber  of  figures  in  the  new  column,  which  are  6 
three,  thus:  2+3=5;  prefix  this  5  to  the  2 
figure  2,  under  the  original  column ;  this  — 
makes  52,  which  is  the  sum  of  the  cokmn.  52 


ADDITION.  13 

RULE. — For  adding  two  or  more  columns,  com- 
mence at  the  right  hand,  or  units'  column;  proceed 
in  the  same  manner  as  in  adding  one  column  ;  after 
the  sum  of  the  first  column  is  obtained,  add  all 
except  the  right  hand  figure  of  this  sum  to  the  second 
column,  adding  the  second  column  the  same  way  you 
added  the  first ;  proceed  in  like  manner  with  all  the 
columns,  always  adding  to  each  successive  column 
the  sum  of  the  column  in  the  next  lower  order,  minus 
the  right  hand  figure. 

N.  B.  The  small  figures  which,  we  place  to  the 
right  of  the  column  when  adding  are  called  integers. 

The  addition  by  integers  or  by  forming  a  new 
column,  as  explained  in  the  preceding  examples 
should  be  used  only  in  adding  very  long  columns 
of  figures,  say  a  long  ledger  column,  where  the  foot- 
ings of  each  column  would  be  two  or  three  hundred, 
in  which  case  it  is  superior  and  much  more  easy 
than  any  other  mode  of  addition ;  but  in  adding 
short  columns  it  would  be  useless  to  form  an  extra 
column,  where  there  is  only,  say,  six  or  eight  fig- 
ures to  be  added.  In  making  short  additions,  the 
following  suggestions  will,  we  trust,  be  of  use  to 
the  accountant  who  seeks  for  information  on  this 
subject. 

In  the  addition  of  several  columns  of  figures, 
where  they  are  only  four  or  five  deep,  or  when 
their  respective  sums  will  range  from  twenty-five 


14         ORTON'S  LIGHTNING  CALCULATOR. 

to  forty,  the  accountant  should  commence  with  the 
unit  column,  adding  the  sum  of  the  first  two  figures 
to  the  sum  of  the  next  two,  and  so  on,  naming  only 
the  results  that  is  the  sum  of  every  two  figures. 

In  the  present  example  in  adding  the  unit  346 
column  instead  of  saying  8  and  4  are  12  and  235 
5  are  17  and  6  are  23,  it  is  better  to  let  the  724 
eye  glide  up  the  column -reading  only,  8,  12,  598 
17,  23 ;  and  still  better,  instead  of  making  a 
separate  addition  for  each  figure,  group  the  figures 
thus :  12  and  11  are  23,  and  proceed  in  like  man- 
ner with  each  column.  For  short  columns  this  is 
a  very  expeditious  way,  and  indeed  to  be  preferred; 
but  for  long  columns,  the  addition  by  integers  is 
the  most  useful,  as  the  mind  is  relieved  at  intervals 
and  the  mental  labor  of  retaining  the  whole  amount, 
as  you  add,  is  avoided,  which  is  very  important  to 
any  person  whose  mind  is  constantly  employed  in 
various  commercial  calculations. 

In  adding  a  long  column,  where  the  figures  are 
of  a  medium  size,  that  is,  as  many  8s  and  9s  as 
there  are  2s  and  3s,  it  is  better  to  add  about  three 
figures  at  a  time,  because  the  eye  will  distinctly  see 
that  many  at  once,  and  the  ingenious  student  will 
in  a  short  time,  if  he  adds  by  integers,  be  able  to 
read  the  amount  of  three  figures  at  a  glance  or  as 
quick,  we  might  say,  as  he  would  read  a  single 
figure. 


ADDITION.  15 

Here  we  begin  to  add  at  the  bottom  of  the     6268 
anit  column  and  add  successively  three  fig-       "• 
ares  at  a  time,  and  place  their  respective       004 
sums,  minus  10,  to  the  right  of  the  last  fig-     554. 
are  added;  if  the  three  figures  do  not  make       62 
10,  add  on  more  figures;  if  the  three  figures       87j» 
make  20  or  more,  only  add  two  of  the  fig-       ^ 
tires.     The  little  figures  that  are  placed  to       ^.4 
the  right  and  left  of  the  column  are  called     877 
integers.     The  integers  in  the  present  ex-       33 
ample,  belonging  to  the  units  column,  are 
4,  4,  5, 4,  6,  which  we  add  together,  making       ^ 

23;  place  down  3  and  add  2  to  the  number  

of  integers,  which  gives  7,  which  we  add  to     803 
the  tens  and  proceed  as  before. 

REASON. — In  the  above  example,  every  time  wo 
placed  down  an  integer  we  discarded  a  ten,  and 
when  we  set  down  the  3  in  the  answer  we  dis- 
carded two  tens;  hence,  we  add  2  on  to  the  num- 
ber of  integers  to  ascertain  how  many  tens  were 
discarded;  there  being  5  integers  it  made  7  tens, 
which  we  now  add  to  the  column  of  tens;  on  the 
same  principle  we  might  add  between  20  and  30, 
always  setting  down  a  figure  before  we  got  to  30; 
then  every  integer  set  down  would  count  for  2  tens, 
being  discarded  in  the  same  way,  it  does  in  the 
present  instance  for  one  ten.  When  we  add  be- 
tween 10  and  20,  and  in  very  long  columns,  it 


16         OETON'S  LIGHTNING  CALCULATOR. 

would  be  much  better  to  go  as  near  30  as  possible, 
and  count  2  tens  for  every  integer  set  down,  in 
which  case  we  would  set  down  about  one -half  ai 
many  integers  as  when  we  write  an  integer  foi 
every  ten  we  discard. 

When  adding  long  columns  in  a  ledger  or  day  * 
book,  and  where  the  accountant  wishes  to  avoid  the 
writing  of  extra  figures  in  the  book,  he  can  place  a 
strip  of  paper  alongside  of  the  column  he  wishes 
to  add,  and  write  the  integers  on  the  paper,  and  i» 
this  way  the  column  can  be  added  as  convenient 
almost  as  if  the  integers  were  written  in  the  book. 

Perhaps,  too,  this  would  be  as  proper  a  time  as 
any  other  to  urge  the  importance  of  another  good 
habit;  I  mean  that  of  making  plain  figures.  Some 
persons  accustom  themselves  to  making  mere 
scrawls,  and  important  blunders  are  often  the  result. 
If  letters  be  badly  made  you  may  judge  from 
such  as  are  known;  but  if  one  figure  be  illegible, 
its  value  can  not  be  inferred  from  the  others.  The 
vexation  of  the  man  who  wrote  for  2  or  3  monkeys, 
and  had  203  sent  him,  was  of  far  less  importance 
than  errors  and  disappointments  sometimes  result- 
ing from  this  inexcusable  practice. 

We  will  now  proceed  to  give  some  methods  of 
proof.  Many  persons  are  fond  of  proving  the  cor- 
rectness of  work,  and  pupils  are  often  instructed 
to  do  so,  for  the  double  purpose  of  giving  them 


ADDITION.  17 

x 

exercise  in  calculation  and  saving  their  teacher  the 
trouble  of  reviewing  their  work. 

There  are  special  modes  of  proof  of  elementary 
operations,  as  by  casting  out  threes  or  nines,  or  by 
changing  the  order  of  the  operation,  as  in  add- 
ing upward  and  then  downward.  In  Addition, 
some  prefer  reviewing  the  work  by  performing  the 
Addition  downward,  rather  than  repeating  the 
ordinary  operation.  This  is  better,  for  if  a  mis- 
take be  inadvertently  made  in  any  calculation, 
and  the  same  routine  be  again  followed,  we  are  very 
liable  to  fall  again  into  the  same  error.  If,  for 
instance,  in  running  up  a  column  of  Addition  you 
should  say  84  and  8  are  93,  you  would  be  liable,  in 
going  over  the  same  again,  in  the  same  way  to 
slide  insensibly  into  a  similar  error ;  but  by  begin- 
ning at  a  different  point  this  is  avoided. 

This  fact  is  one  of  the  strongest  objections  to 
the  plan  of  cutting  off  the  upper  line  and  adding 
it  to  the  sum  of  the  rest,  and  hence  some  cut  off 
the  lower  line  by  which  the  spell  <s  broken.  The 
most  thoughtless  can  not  fail  to  see  that  adding  a 
line  to  the  sum  of  the  rest,  is  the  same  as  adding  it 
in  with  the  rest. 

The  mode  off  proof  by  casting  out  the  nines 
and  threes  will  be  fully  explained  in  a  following 
chapter. 

A  very  excellent  mode  of  avoiding  error  in  add- 


1  j         ORTON'S  LIGHTNING  CALCULATOR. 

f  fig  long  columns  is  to  set  down  the  result  of  each 
column  on  some  waste  spot,  observing  to  place  the 
numbers  successively  a  place  further  to  the  left 
each  time,  as  in  putting  down  the  product  figures 
in  multiplication  j  and  afterward  add  up  the 
amount.  In  this  way  if  the  operator  lose  his 
count,  he  is  not  compelled  to  go  back  to  units,  but 
only  to  the  foot  of  the  column  on  which  he  is  op- 
erating. It  is  also  true  that  the  brisk  accountant, 
who  thinks  on  what  he  is  doing,  is  less  liable  to 
err  than  the  dilatory  one  who  allows  his  mind  to 
wander.  Practice  too  will  enable  a  person  to  read 
amounts  without  naming  each  figure,  thus  instead 
of  saying  8  and  6  are  14,  and  7  are  21  and  5  are  26, 
it  is  better  to  let  the  eye  glide  up  the  column,  read- 
ing only  8,  14,  21,  26,  etc.;  and,  still  further,  it  is 
quite  practicable  to  accustom  one's  self  to  group  87 
the  figures  in  adding,  and  thus  proceed  very  rap-  23 
idly.  Thus  in  adding  the  units'  column,  instead  45 
of  adding  a  figure  at  a  time,  we  see  at  a  glance  62 
that  4  and  2  are  6,  and  that  5  and  3  are  8,  then  24 
6  and  8  are  14;  we  may  then,  if  expert,  add  — 
constantly  the  sum  of  two  or  three  figures  at  a  time, 
and  with  practice  this  will  be  found  highly  advan- 
tageous in  long  columns  of  figures ;  or  two  or  three 
columns  may  be  added  at  a  time,  as  the  practiced 
eye  will  see  that  24  and  62  are  86  almost  as  readily 
as  that  4  and  2  are  6. 


ADDITION. 


T^abkdrs  trtil  ind  tlie  following  mode  of  match- 
tog  lines  toi  beginners  very  convenient,  as  they  can 
inspect  them  at  a  glance : 


inspect  them  at  a  glance : 

•^v 


Add  7654384 
8786286 
3408698 
2345615 
1213713 


23408696 

\ 
In  placing  the  above  the  lines  are  matched  in 

pairs,  the  digits  constantly  making  9.  In  the 
above,  the  first  and  fourth,  second  and  fifth  are 
matched;  and  the  middle  is  the  key  line,  the  Result 
being  just  like  it,  except  the  units'  place,  which  is 
as  many  less  than  the  units  in  the  key  line  as  there 
are  pairs  of  lines;  and  a  similar  number  will  oc- 
cupy the  extreme  left.  Though  sometimes  used  aa 
a  puzzle,  it  is  chiefly  useful  in  teaching  learners ; 
and  as  the  location  of  the  key  line  may  be  changed 
in  each  successive  example,  if  necessary,  the  arti* 
fice  could  not  be  detected.  The  number  of  lines 
is  necessarily  odd. 

If  the  student  will  practice  the  addition  of  long  col- 
ams  by  integers,  he  will  in  a  short  time  becomo  so 
2 


20  oaTON  &  FULLER'S  ARITHMETIC. 

proficient  in  its  application  that  the  forming  of 
new  columns  will  be  unnecessary;  and  he  will  only 
add  the  number  of  units  in  excess  of  the  tens  in 
each  column.  In  adding  up  dollars  and  cents  in 
a  memorandum,  ledger,  or  day-book,  the  beginner 
should  add  with  red  ink,  so  as  to  determine  readily 
the  new  column  from  the  original. 


MULTIPLICATION. 

MULTIPLICATION,  in  its  most  general  sense,  is  a 
series  of  additions  of  the  same  number ;  therefore, 
in  multiplication,  a  number  is  repeated  a  certain 
number  of  times,  and  the  result  thus  obtained  is 
called  the  product.  When  the  multiplicand  and  the 
multiplier  are  each  composed  of  only  two  figure?, 
to  ascertain  the  product  we  have  the  following 

RULE. — Set  down  the  smaller  factor  under  the 
larger,  units  under  units,  tens  under  tens.  Begin 
with  the  unit  figure  of  the  multiplier ',  multiply  by  it, 
first  the  units  of  the  'multiplicand,  setting  the  units 
of  the  product,  and  reserving  the  tms  to  be  added 
to  the  next  product;  now  multiply  the  tens  of  the 
multiplicand  by  the  unit  figure  of  the  multiplier, 
znd  the  units  of  the  multiplicand  by  tons  figure  of 


MULTIPLICATION.  21 

the  multiplier;  add  these  two  products  together,  set 
Una  down  the  units  of  their  sum,  and  reserving  the 
lens  to  be  added  to  the  next  product ;  now  multiply 
(he  tens  of  the  multiplicand  by  the  tens  figure  of  the 
multiplier,  and  set  down  the  whole  amount.  Th'Ji 
will  be  the  complete  product. 

Remark. — Always  add  in  the  tens  that  are  rt 
served  as  soon  as  you  form  the  first  product. 

EXAMPLE  1.— EXPLANATION. 

1.  Multiply  the  units  of  the  multiplicand       24 
by  the  unit  figure  of  the  multiplier,  thus :       31 

IX^  is  4 ;  set  the  4  down  as  in  example.     • 

2.  Multiply  the  tens  in  the  multiplicand  by  744 
the  unit  figure  in  the  multiplier,  and  the  units  in 
the  multiplicand  by  the  tens  figure  in  the  multi- 
plier, thus  :  1X2  is  2;  3X4  are  12,  add  these  two 
products  together,  2-J-12  are  14,  set  the  4  down 
as  in  example,  and  reserve  the  1  to  be  added  to  the 
next  product.  3.  Multiply  the  tens  in  the  multi- 
plicand by  the  tens  figure  in  the  multiplier,  and 
add  in  the  tens  that  were  reserved,  thus :  3x2  are 
6,  and  6+1=7 ;  now  set  down  the  whole 
amount,  which  is  7. 

EXAMPLE  2.— EXPLANATION. 

1.  Multiply  units  by  units,  thus:  4x3         53 
are  12,  set  down  the  2  and  reserve  the  1  to         84 

carry.    2.  Multiply  tens  by  units,  and  units     

by  tens,  and  add  in  the  one  to  carry  on  the    4i52 


22  OETON  &  FULLER'S  ARITHMETIC. 

first  product,  then  add  these  two  products  together, 
ihus:  4X5  are  20+1  are  21,  and  8X3  are  24, 
and  21+24  are  45,  set  down  the  5  and  reserve 
£he  4  to  carry  to  the  next  product.  3.  Multiply 
tens  by  tens,  and  add  in  what  was  reserved  to  carry, 
thus  :  8X5  are  40+4  are  44,  now  set  down  the 
whole  amount,  which  is  44. 

EXAMPLE  3.— EXPLANATION. 

5X^  are  15,  set  down  the  5  and  carry  the  43 
I  to  the  next  product;  5X4  are  20=1  25 

are  21;  2x3  are  6,  21+6  are  27,  set  down     

the  7  and  carry  the  2;  2x4  are  8+2  are  1075 
10  ;  now  set  down  the  whole  amount. 

When  the  multiplicand  is  composed  of  three  fig- 
ares,  and  there  are  only  two  figures  in  the  multi- 
plier, we  obtain  the  product  by  the  following 

RULE. — Set  down  the  smaller  factor  under  the 
larger,  units  under  units,  tens  under  tens  ;  now  mul* 
tiply  the  first  upper  figure  by  the  unit  figure  of  the 
multiplier,  setting  down  the  units  of  the  product,  and 
reserving  the  tens  to  be  added  to  the  next  product; 
now  multiply  the  second  upper  by  units,  and  the  first 
upper  by  tens,  add  these  two  products  together,  set- 
ting down  the  units  figure  of  their  sum,  and  reserv- 
ing the  tens  to  carry,  as  before;  now  multiply  the 
ihird  upper  by  units,  and  the  second  upper  by  tens, 
add  these  two  products  together,  setting  down  the 
units  figure  of  their  sum,  and  reserving  the  tens  to 


MULTIPLICATION.  2? 

,  as  usual ;  now  multiply  the  third  upper  ly 
tens,  add  in  the  reserved  figure,  if  there  is  one,  and 
set  down  the  whole  amount.  This  will  be  the  com" 
plete  product. 

Remark. — -One  of  the  principal  errors  with  the 
beginner,  in  this  system  of  multiplication,  is 
neglecting  to  add  in  the  reserved  figure.  The  stu- 
dent must  bear  in  mind  that  the  reserved  figure  is 
added  on  to  the  first  product  obtained  after  the  set- 
ting down  of  a  figure  in  the  complete  product. 

EXAMPLE  1.— EXPLANATION. 

Multiply  first  upper  by  units,  5X3  are  123 
15,  set  down  the  5,  reserve  the  1  to  carry  45 

to  the  next  product;  now  multiply  second     

upper  by  units  and  first  upper  by  tens,  5x2  5535 
are  10+1  are  11,4x3  are  12,  add  these 
products  together;  ll-f-12  are  23,  set  down  the  3, 
reserve  the  2  to  carry ;  now  multiply  third  upper 
by  units,  and  second  upper  by  tens,  add  these  two 
products  together,  always  adding  on  the  reserved 
figure  to  the  first  product;  5x1  are  5+2  are  7, 
4X2  are  8,  and  7+8  are  15,  set  down  the  5,  re- 
Sfrrv*  the  1 ;  now  multiply  third  upper  by  tens, 
and  set  down  the  whole  amount;  4X1  are  4+1  are 
5,  set  down  the  5.  This  will  give  the  complete 
product. 


24         ORTON'S  LIGHTNING  CALCULATOR. 

Multiply  32  by  45  in  a  single  line. 

Here  we  multiply  5x2  and  set  down  and  32 
carry  as  usual ;  then  to  what  you  carry  add  45 

5X3  and  4X  2,  which  gives  24;  set  down     

4  and  carry  2  to  4x3,  which  gives  14  and     1440 
completes  the  product. 

Multiply  123  by  456  in  a  single  line. 

Here  the  first  and  second  places  are  123 
found  as  before;  for  the  third,  add  6Xl,  456 

5X2,  4x3,  with  the  2  you  had  to  carry,  

making  30  ;  set  down  0  and  carry  3 ;  then  56088 
drop  the  units*  place  and  multiply  the 
hundreds  and  tens  crosswise,  as  you  did  the  tens 
and  units,  and  you  find  the  thousand  figure ;  then, 
dropping  both  units  and  tens,  multiply  the  4Xl» 
adding  the  1  you  carried,  and  you  have  5,  which 
completes  the  product.  The  same  principle  may 
be  extended  to  any  number  of  places;  but  let  each 
step  be  made  perfectly  familiar  before  advancing 
to  another.  Begin  with  two  places,  then  take  three, 
then  four,  but  always  practicing  some  time  on  each, 
number,  for  any  hesitation  as  you  progress  will, 
confuse  you. 

N.  B.  The  following  mode  of  multiplying  num- 
bers will  only  apply  where  the  sum  of  the  two  last 
or  unit  figures  equal  ten,  and  the  other  figures  in 
both  factors  are  the  same. 


MULTIPLICATION  26 

-    V 

CONTRACTIONS  IN  MULTIPLICATION. 

To  multiply  when  the  unit  figures  added  equal 
(10)  and  the  tens  are  alike  as  72  &#  78,  &c. 

1st.  Multiply  the  units  and  set  down  the  result. 

2d.  Add  1  to  either  number  in  tens  place  and 
multiply  by  the  other,  and  you  have  the  complete 
product. 

EXAMPLE  FIRST — PROCESS. 

Here  because  the  sum  of  the  units  4  and  6  86 
are  ten  and  the  tens  are  alike  ;  we  simply  say  84 
4  times  6  are  24,  and  set  down  both  figures  of 
the  product ;  then  because  4  and  6  make  ten  we 
add  1  to  8,  making  9,  and  9  times  8  are  72,  which 
completes  the  product. 

NOTE. — If  the  product  of  units  do  not  contain  ten  the  place 
of  tens  must  be  filled  with  a  cipher 

The  above  rule  is  useful  in  examples  like  the  fol- 

2.  What  will  93  acres  of  land  cost  at  97  dollars 
lowing : 

per  acre?  Ans.  $9021. 

3.  What  will  89  pounds  of  tea  cost  at  81  cents 
per  pound  ?  Ans.  $72.09. 

In  the  above  the  product  of$by\  did  not  amount 
to  ten,  therefore  0  is  placed  in  tens  place. 

4.  Multiply  998  by  992.  '  Ans.  990016. 
In  the  above,  because  2  and  8  are  10,  we  add  1  to 

99,  making  100;  then  100  times  99  are  9900. 


26         OKTON'S  LIGHTNING  CALCULATOR. 

EXAMPLE  EIGHTEENTH. 

Multiply  79  by  71  in  a  single  line. 

Here  we  multiply  lX^  and  set  down  the  79 
result,  then  we  multiply  the  7  in  the  mul-  71 
tiplicand,  increased  by  1  by  the  7  in  the  — — 
multiplier,  7x8,  which  gives  56  and  com-  5609 
pletes  the  product. 

EXAMPLE   NINETEENTH. 

Multiply  197  by  193  in  a  single  line. 

Here  we  multiply  3x7  and  set  down  the  197 
result,  then  we  multiply  the  19  in  the  193 

multiplicand,  increased  by  1  by  the  19  in 

the  multiplier,  19x20,  which  gives  380  38021 
and  completes  the  product. 

EXAMPLE  TWENTIETH. 

Multiply  996  by  994  in  a  single  line. 

Here  we  multiply  4x6  and  set  down  996 

the  result,  then  we  multiply  the  99  in  994 

the  multiplicand,  increased  by  1  by  the  *- 

99  in  the  multiplier,  99x100,  which  990024 
gives  9900  and  completes  the  product. 

EXAMPLE  TWENTY-FIRST. 

Multiply  1208  by  1202  in  a  single  line. 

Here  we  multiply  2x8  and  set  down  1208 
the  result,  then  we  multiply  the  120  in  1202 

the  multiplicand,  increased  by  1  by  the 

120  in  the  multiplier,  120X121,  which  1452016 
gives  14520  and  completes  the  product. 


MULTIPLICATION.  27 

CURIOUS  AND  USEFUL  CONTRACTIONS. 

f  o  multiply  any  number,  of  two  figures,  by  11, 
KpLE. —  Write  the  sum  of  the  figures  between  them. 

1.  Multiply  45  by  11.  Ans.  495 
Here  4  and  5  are  9,  which  write  between  4  &  5, 

2.  Multiply  34  by  11.  Ans.  374. 
N.  B.  When  the  sum  of  the  two  figures  is  over 

9,  increase  the  left-hand  figure  by  the  1  to  carry. 

3.  Multiply  87  by  11.  Ana.  957. 

To  square  any  number  of  9s  instantaneously, 
and  without  multiplying, 

RULE. —  Write  down  as  many  9s  less  one  as  there 
are  9s  in  the  given  number,  an  8,  as  many  Os  at 
9s,  and  a  1. 

4.  What  is  the  square  of  9999  ?    Ans.  99980001. 
EXPLANATION. — We  have  four  9s  in  the  given 

number,  so  we  write   down  three  9s,  then  an  8, 
then  three  Os,  and  a  1. 

5.  Square  999999.  Ans.  999998000001. 

To  square  any  number  ending  in  5, 

RULE. —  Omit  the  5  and  multiply  the  number,  as 

it  will  then  stand  by  the  next  higher  number,  and 

annex  25  to  the  product. 

6.  What  is  the  square  of  75  ?  Ans.  5625. 
EXPLANATION. — We  simply  say,  7  times  8  arc 

56,  to  which  we  annex  25. 

7.  What  is  the  square  of  95?  Ans.  9025 


28  ORTON'S    LIGHTNING    CALCULATOR. 

Mental  Operations  in  Fractions. 

To  square  any  number  containing  |,  as  6^,  9^> 

RULE. — Multiply  the  whole  number  by  the  next 
higher  whole  number,  and  annex  ^  to  the  product. 

Ex.  1.  What  is  the  square  of  7£?     Ans.  56£. 

We  simply  say,  7  times  8  are  56,  to  which  we 
add  I. 

2.  What  will  91  Ibs.  beef  cost  at  9^-  cts.  a  lb.? 

3.  What  will  12^  yds.  tape  cost  at  12^ cts.  a  yd.? 

4.  What  will  54-  Ibs.  nails  cost  at  5^  cts.  a  lb.  ? 

5.  What  will  ll|  yds.  tape  cost  at  11-j-  cts.  a  yd.  ? 

6.  What  will  19£  bu.  bran  cost  at  19|  cts.  a  bu.? 
REASON. — We  multiply  the  whole  number  by 

the  next  higher  whole  number,  because  half  of  any 
number  taken  twice  and  added  to  its  square  is  the 
same  as  to  multiply  the  given  number  by  ONE  more 
than  itself.  The  same  principle  will  multiply  any 
two  like  numbers  together,  when  the  sum  of  the 
fractions  is  ONE,  as  8£  by  8|,  or  llf  by  llf,  etc 
It  is  obvious  that  to  multiply  any  number  by  any 
two  fractions  whose  sum  is  ONE,  that  the  sum  of  the 
products  must  be  the  original  number,  and  adding  the 
number  to  its  square  is  simply  to  multiply  it  by 
ONE  more  than  itself;  for  instance,  to  multiply  7-J 
by  7|,  we  simply  say,  7  times  8  are  56,  and  then, 
to  complete  the  multiplication,  we  add,  of  course, 
the  product  of  the  fractions  (|  times  £  are  A)> 
making  56T3g  the  answer. 


MULTIPLICATION.  29 

Where  the  sum  of  the  Fractions  is  ONE. 

To  multiply  any  two  like  numbers  together  when 
the  sum  of  the  fractions  is  ONE. 

RULE.  —  Multiply  the  whole  number  by  the  next 
higher  whole  number;  after  which,  add  the  product 
of  the  fractions. 

N.  B.  In  the  following  examples,  the  product  of 
the  fractions  are  obtained  first  for  convenience. 

PRACTICAL  EXAMPLES  FOR  BUSINESS  MEN. 

Multiply  3|  by  3^  in  a  single  line. 

Here  we  multiply  ^X|j  which  gives  T35,  3| 
and  set  down  the  result  ;  then  we  multiply  3^ 
the  3  in  the  multiplicand,  increased  by  - 
unity,  by  the  3  in  the  multiplier,  3X^,  12TV 
which  gives  12  and  completes  the  product. 

Multiply  7f  by  7f  in  a  single  line. 

Here  we  multiply  f  Xf>  which  gives  5\,  7f 
*nd  set  down  the  result;  then  we  multiply  7^ 
the  7  in  the  multiplicand,  increased  by  unity,  - 
oy  the  7  in  the  multiplier,  7x$,  which  gives 
66,  and  completes  the  product. 

Multiply  11-|  by  llf  in  a  single  line. 


Here  we  multiply  -|  X£,  which  gives  f,  and  11  J- 
set  down  the  result;  then  we  multiply  the  11  llf 
in  the  multiplicand,  increased  by  unity,  by  -- 
the  11  in  the  multiplier,  11X.12,  which  gives  132| 
132,  and  completes  the  product. 


3)         ORTON'S  LIGHTNING  CALCULATOR. 

EXAMPLE  THIRTY-THIRD. 

Multiply  16f  by  16  J  in  a  single  line. 

Here  we  multiply  Jxf  which  gives  1,  and  16$ 
set  down  the  result,  then  we  multiply  the  16 J 

16    in    the     multiplicand,    increased     by    

unity  by  the  16  in  the  multiplier,  16X17,     272§ 
which  gives  272  and  completes  the  product. 
EXAMPLE  THIRTY-FOURTH. 

Multiply  29J  by  29J  in  a  single  line. 

Here  we  multiply  JX^  which  gives  J,       29 J 
and  set  down  the  result,  then  we  multiply 
the  29  in  the  multiplicand,  increased  by 
unity   by  the   29  in  the  multiplier,   29  X     870J 
30,  which  gives  870  and  completes  the  pro- 
duct. 

EXAMPLE  THIRTY-FIFTH. 

Multiply  999f  by  999f  in  a  single  line. 

Here  we  multiply  fXf3  which  gives  999f 

J|,  and  set  down  the  result,  then  we  999|- 

multiply  the  999  in  the  multiplicand, 

increased  by  unity  by  the  999  in  the     999000£f 
multiplier,    999X1000,    which    gives 
999000  and  completes  the  product. 

NOTE. — The  system  of  multiplication  introduced 
in  the  preceding  examples,  applies  to  all  numbers, 
\Vhcre  the  sum  of  the  fractions  is  onej  and  the  whole 
numbers  are  alike,  or  differ  by  one,  the  learner  is 
requested  to  study  well  these  useful  properties  of 
numbers. 


ORTON  S    LIGHTNING   CALCULATOR.  81 

WJiere  the  sum  of  the  Fractions  is  ONE. 

To  multiply  any  two  numbers  whose  difference 
ip  one,  and  the  sum  of  the  fractions  is  one, 

RULE.  —  Multiply  the  larger  number,  increased  by 
ONE,  by  the  smaller  number;  then  square  the  frac- 
tion of  the  larger  number  ,  and  subtract  its  square 
from  ONE. 

PRACTICAL  EXAMPLES  FOR  BUSINESS  MEN. 

1.  What  will  9£  Ibs.  sugar  cost  at  8|  cts.  a  lb.? 
Here  we  multiply  9,  increased  by  1,  by  8,         91 

thus,  8x10  are  80,  and  set  down  the  result;  g3 
then  from  1  we  subtract  the  square  of  -J,  -- 
thus,  £  squared  is  T^-,  and  1  less  -^  is  |-|.  &OJ-J 

2.  What  wih  8|  bu.  coal  cost  at  7|  cts.  a  bu.? 
Here  we  multiply  8,  increased  by  1,  by  S-f 

7,  thus,  7  times  9  are  63,  and  set  down  the  7^ 

result;  then  from  1  we  subtract  the  square  - 
of  -f  ,  thus,  -|  squared  is  f  ,  and  1,  less  •£,  is  f  .  "S 

3.  What  will  11^  bu.  seed  cost  at  $10f|  a  bu.? 
Here  we  multiply  11,  increased  by  1,  by 

10,  thus,  10   times   12  are   120,   and  set     11i\ 
down  the  result;  then  from  1  we  subtract     "^T5 
the  square  of  ^  thus,  ^  squared  is  Tfo,  120i«5 
and  1  less          is 


4.  How  many  square  inches  in  a  floor  99|^  in- 
wide  and  98|  in.  long?  Ans.  9800&J. 


32        ORTON'S  LIGHTNING  OALCILATOE. 

METHOD  OF  OPERATION. 
EXAMPLE  FIRST. 

Multiply  6J  by  6J-  in  a  single  line. 

Here  we  add  6J-|- J,  which  gives  6J ;  this  6] 
multiplied  by  the  6  in  the  multiplier,  6J 

6X6*2,  gives  39,  to  which  we  add  the  pro-  

duct  of  the  fractions,  thus  JXi  gives  y'g,  added 
39_ig  to  39  completes  the  product. 

EXAMPLE  SECOND. 

Multiply  11 J  by  llf  in  a  single  line.    . 

Here  we  would  add  llj-j-f,  which  gives  11 J 
12;  this  multiplied  by  the  11  in  the  multi-  31f 

plier  gives  132,  to  which  we  add  the  product  

of  the  fractions,  thus  f Xj  gives  T35,  which  13278g 
added  to  132  completes  the  product. 

EXAMPLE  THIRD. 

Multiply  12J  by  12f  in  a  single  line. 

Here  we  add  12J+f,  which  gives  13J;  12J 
this  multiplied  by  the  12  in  the  multiplier,  12| 

12X13^,  gives  159,  to  which  add  the  pro 

duct  of  the  fractions,  thus  |Xi  gives  f,  159f 
which  added  to  159  completes  the  product. 


ORTON'S  LIGHTNINGI  CALCULATOR.         33 

Where  the  Fractions  have  a  Like  Denominator. 

To  multiply  any  two  like  numbers  together,  each 
of  which  has  a  fraction  with  a  like  denominator,  as 
4f  by  4|,  or  11^  by  11 1,  or  lOf  by  lOi,  etc. 

RULE. — Add  to  the  multiplicand  the  fraction  of 
the  multiplier,  and  multiply  this  sum  by  the  whole 
number;  after  which,  add  the  product  of  the  fractions 

PRACTICAL  EXAMPLES  FOR  BUSINESS  MEN. 

N.  B.  In  the  following  example,  the  sum  of  the  frac- 
tions is  ONE. 

1.  What  will  9|  Ibs.  beef  cost  at  9^  cts.  a  lb.? 

The  sum  of  9|-  and  ^  is  ten,  so  we  simply  9-| 
say,  9  times  10  are  90;  then  we  add  the  * 
product  of  the  fractions,  |  times  |  are  T3^-.  90^ 

N.  B.  In  the  following  example,  the  sum  of  the  frao 
tions  \s  less  than  ONE. 

2    What  will  8J  yds.  tape  cost  at  8|  cts.  a  yd.  ? 

The  sum  of  8|  and  f  is  8|,  so  we  simply       8£ 
say,  8  times  8f  are  70;    then  we  add  the 
product  of  the  fractions,  -f  times  ^  are  ^  or  -J-.     70| 

N.  B.  In  the  following  example,  the  sum  of  the  frac- 
tions is  greater  than  ONE. 

3.  What  will  4|  yds.  cloth  cost  at  $4|  a  yd.? 

The  sum  of  4|  and  |-  is  5£,  so  we  simply  4| 
say,  4  times  5^  are  21;  then  we  add  the  » 
product  of  the  fractions,  J  times  |  are  |J.  21§-| 

N.  B.  "Where  the  fractions  have  different  denominatoru, 
reduce  them  to  a  common  denominator. 
3 


34         ORTON'S  LIGHTNING  CALCULATOR. 

Rapid  Process  of  Multiplying  Mixed  Numbers. 

A  valuable  and  useful  rule  for  the  accountant  in 
the  practical  calculations  of  the  counting-room. 

To  multiply  any  two  numbers  together,  each  of 
which  involves  the  fraction  -i,  as  7£  by  9£,  etc., 

RTJLE. —  To  the  product  of  the  whole  numbers  add 
"kalf  their  sum  plus  £. 

EXAMPLES  FOR  MENTAL  OPERATIONS. 

1.  What  will  3£doz.  eggs  cost  at  7|  cts.  a  doz.? 
Here  the  sum  of  7  and  3  is  10,  and  half  this     31 

sum  is  5,  so  we  simply  say,  7  times  3  are  21     7-J 

and  5  are  26,  to  which  we  add  4. 

264 
N.  B.  If  the  sum  be  an  odd  number,  call  it  one  less 

tc  make  it  even,  and  in  such  cases  the  fraction  must  be  }. 

2.  What  will  11|  Ibs.  cheese  cost  at  9£  cts.  a  lb.? 

3.  What  will  8£  yds.  tape  cost  at  15£  cts.  a  yd.? 

4.  What  will  7£  Ibs.  rice  cost  at  131  cts.  a  lb.? 

5.  What  will  lOj-  bu.  coal  cost  at  12£  cts.  a  bu.? 
REASON. — In  explaining  the  above  rule,  we  add 

half  their  sum  because  half  of  either  number  added 
to  half  the  other  would  be  half  their  sum,  and  we 
add  -^  because  %  by  %  is  \.  The  same  principle 
will  multiply  any  two  numbers  together,  each  of 
which  has  the  same  fraction;  for  instance,  if  the 
fraction  was  -J-,  we  would  add  one-third  their  sum ; 
if  f,  we  would  add  three-fourths  their  sum,  etc.; 
and  then,  to  complete  the  multiplication,  we  would 
add,  of  course,  the  product  of  the  fractions. 


MULTIPLICATION.  35 

GENERAL  RULE 
For  multiplying  any  two  numbers  together,  each 

of  which  involves  the  same  fraction. 

To  the  product  of  the  whole  numbers,  add  tht 
product  of  their  sum  by  either  fraction;  after  which 
add  the  product  of  their  fractions. 

EXAMPLES  FOR  MENTAL  OPERATIONS. 

1.  What  will  llf  Ibs.  rice  cost  at  9f  cts.  a  lb.? 
Here  the  sum  of  9  and  11  is  20.  and  three-     ]\z 

fourths  of  this  sum  is  15,  so  we  simply  say,       9| 

9  times  11  are  99  and  15  are  114,  to  which  

we  add  the  product  of  the  fractions  (3^-).  ra 

2.  What  will  7f  doz.  eggs  cost  at  8|  cts.  a  doz.  ? 

3.  What  will  6f  bu.  coal  cost  at  6|  cts.  a  bu.  ? 

4.  What  will  45f  bu.  seed  cost  at  3f  dol.  a  bu.? 

5.  What  will  3|  yds.  cloth  cost  at  5f  dol.  a  yd.  ? 

6.  What  will  17f  ft.  boards  cost  at  13f  cts  a  ft.? 

7.  What  will  18|  Ibs.  butter  cost  at  18|  cts.  a  lb.  ? 
N.  B.  If  the  product  of  the  sum  by  either  frac 

tion  is  a  whole  number  with  a  fraction,  it  is  bettei 
to  reserve  the  fraction  until  we  are  through  with 
the  whole  numbers,  and  then  add  it  to  the  product 
of  the  fractions;  for  instance,  to  multiply  3^  by  7J, 
we  find  the  sum  of  7  and  3,  which  is  10,  and  one 
fourth  of  this  sum  is  2.1 ;  setting  the  |  down  in 
some  waste  spot,  we  simply  say,  7  times  3  are  21 
and  2  are  23 ;  then,  adding  the  \  to  the  product  of 
the  fractions  (y6),  gives  T\,  making  23^,  Ans 


36         ORTON'S  LIGHTNING  CALCULATOR. 

Rapid  Process  of  Multiplying  all  Mixed  Numberi. 
N.  B.  Let  the  student  ren) ember  thai  this  is  a 
goneral  and  universal  rule. 

GENELAL  RULE. 

To  multiply  any  two  mixed  numbers  together, 

1st.  Multiply  the  whole  numbers  together. 

2d.    Multiply  the  upper  digit  by  the  lower  fraction. 

3d.    Multiply  the  lower  digit  by  the  upper  fraction. 

4th.  Multiply  the  fractions  together. 

5th.  Add  these  FOUR  products  together. 
N.  B.  This  rule  is  so  simple,  so  useful,  and  so  true  that 
every  banker,  broker,  merchant,  and  clerk  should  post  it 
ap  for  reference  and  use. 

PRACTICAL  EXAMPLES  FOR  BUSINESS  ME3 
N.  B.  The  following  method  is  recommended  to  begin- 
ners : 

EXAMPLE.— Multiply  12f  by  9|.  12| 

1st.  We    multiply   the    whole   numbers.       9f 
2d.    Multiply  12  by  -|  and  write  it  down.  jQg 
3d.    Multiply    9  by  f  and  write  it  down.       9 
4th.  Multiply    f  by  -|  and  write  it  down. 
5th.  Add  these  four  products   together,  _     T2 
and  we  have  the  complete  result.       123T6Tj- 
1ST.  B.  When  the  student  has  become  familiar 
with  the  above  process,  it  is  better  to  do  the  inter- 
mediate WDrk  in  the  head,  and,  instead  of  setting 
down  the  partial  products,  add  them  in  the  mind 
as  you  pass  along,  and  thus  proceed  very  rapidly. 


MULTIPLICATION.  37 

Multiply  8^-  by  10£. 

Here  we  simply  say  10  times  8  are  80  8| 

and  |  of  8  is  2,  making  82,  and  |  of  10  is  10j 

23  which  makes  84;  then  £  times  £  is  -fa,     

•flaking  84^  the  answer.  81^ 

PRACTICAL  BUSINESS  METHOD 
For  Multiplying  all  Mixed  Numbers. 

Merchants,  grocers,  and  business  men  generally, 
in  multiplying  the  mixed  numbers  that  arise  in 
the  practical  calculations  of  their  business,  only 
care  about  having  the  answer  correct  tG  the  near- 
est cent;  that  is,  they  disregard  the  fraction. 
When  it  is  a  half  cent  or  more,  they  call  it  an- 
other cent ;  if  less  than  half  a  cent,  they  drop  it. 
And  the  object  of  the  following  rule  is  to  show  the 
business  man  the  easiest  and  most  rapid  process  of 
finding  the  product  to  the  nearest  unit  of  any  two 
numbers,  one  or  both  of  which  involves  a  fraction. 
GENERAL  RULE. 

To  multiply  any  two  numbers  to  the  nearest  unit, 

1st.  Multiply  the  whole  number  in  the  multiplicand 
by  the  fraction  in  the  multiplier  to  the  nearest  unit. 

2d.  Multiply  the  whole  number  in  the  multiplier  by 
the  fraction  in  the  multiplicand  to  the  nearest  unit 

3d.  Multiply  the  whole  numbers  together  and  add 
the  three  products  in  your  mind  as  you  proceed. 

N.  B.  In  actual  business  the  work  can  generally   be 
done  mentally  for  on?y  easy  fractions  occur  in  business. 


38         ORTON'S  LIGHTNING  CALCULATOR. 

N  B.  This  rule  is  so  simple  and  so  true,  according  to 
all  business  usage,  that  every  accountant  should  make 
himself  perfectly  familiar  with  its  application.  There 
being  no  such  thing  as  a  fractian  to  add  in,  there  is 
scarcely  any  liability  to  error  or  mistake.  By  no  other 
arithmetical  process  can  the  result  be  obtained  by  so  few 
rfgures. 

EXAMPLES  FOR  MENTAL  OPERATION. 
EXAMPLE    FIRST. 

Multiply  11^-  by  8|  by  business  method.      11 1 
Here  ^  of  1 1  to  the  nearest  unit  is  3,  and  ^  of      8 J 

8  to  the  nearest  unit  is  3,  making  6,  so  we  sim-     » 

ply  say,  8  times  11  are  88  and  6  are  94,  Ans.     94 

REASON. — J  of  11  is  nearer  3  than  2,  and  J  of  8  is  nearer 
3  than  2.  Make  the  nearest  whole  number  the  quotient. 

EXAMPLE    SECOND. 

Multiply  7f  by  9|  by  business  method. 

Here  |  of  7  to  the  nearest  unit  is  3,  and  J  7-f 

of  9  to  the  nearest  unit  is  7 ;  then  3  plus  7  9$ 

is  10,  so  we  simply  say,  9  times  7  are  63  and  

10  are  73,  Ans.  73 

EXAMPLE   THIRD. 

Multiply  23i  by  19£  by  business  method. 

Here  ^  of  23  to  the  nearest  unit  is  6,  and  23£ 
£  of  19  to  the  nearest  unit  is  6  ;  then  6  plus  19J 

6  is  12,  so  we  simply  say,  19  times  23  are  

437  and  12  are  449,  Ans. 

N.  B.  In  multiplying  the  whole  numbers  together  al 
ways  use  the  single-line  method. 


MULTIPILCATION.  39 

EXAMPLE    FOURTH. 

Multiply  128f  by  25  by  business  method. 
Here  f  of  25  to  the  nearest  unit  is  17,  so     128  § 
we  simply  say,  25  times  128  are  3200  and ^ 

17  are  3217,  the  answer.  3217 

PRACTICAL  EXAMPLES  FOR  BUSINESS  MEN. 

1.  What  is  the  cost  of  17£  Ibs.  sugar  at  18|  cts, 
per  lb.? 

Here  |  of  17  to  the  nearest  unit  is  13,  17^ 
and  |  of  18;  is  9  13  plus  9  is  22,  so  we  18| 
simply  say.  18  times  17  are  306  and  22  are  

Qi'l  9Q 

328,  the  answer. 

2.  What  is  the  cost  of  11  Ibs.  5  oz.  of  butter  at 
33 J  cts.  per  lb.? 

Here  -J  of  11  to  the  nearest  unit  is  4,       HT$ 
and  ^  of  33  to  the  nearest  unit  is  10  ;       33} 
then  4  plus  10  is  14,  so  we  simply  say,  33 
times  11  are  363,  and  14  are  377,  Ans. 

3.  What  is  the  cost  of  17  doz.  and  9  eggs  at 
12^  cts.  per  doz.? 

Here  %  of  17  to  the  nearest  unit  is  9,       l?^ 
and  ^  of  12  is  9 ;  then  nine  plus  9  is  18,       lf>£ 
BO  we  simply  say,  12  times  17  are  204  and  

18  are  222,  the  answer. 

4.  What  will  be  the  cost  of  15|  yds.  calico  at 
12J  cts.  per  yd.?  Ans.  $1.97. 

N.  B.  To  mul  tiply  by  aliquot  parts  of  100,  see  page  44 
D 


40         ORTON'S  LIGHTNING  CALCULATOR. 


RAPID  PKOCESS  OF  MARKING  GOODS. 

A  VALUABLE  HINT  TO  MERCHANTS  ANP  ALL  RETAIL  DEALERS 
IN  FOREIGN  AND  DOMESTIC  DRY  GOODS. 

RETAIL  merchants,  in  buying  goods  by  whole- 
sale, buy  a  great  many  articles  by  the  dozen,  such 
as  boots  and  shoes,  hats  and  caps,  and  notions  of 
various  kinds.  Now,  the  merchant,  in  buying,  for 
instance,  a  dozen  hats,  knows  exactly  what  one  of 
those  hats  will  retail  for  in  the  market  where  he 
deals ;  and,  unless  he  is  a  good  accountant,  it  will 
often  take  him  some  time  to  determine  whether  he 
can  afford  to  purchase  the  dozen  hats  and  make  a 
living  profit  in  selling  them  by  the  single  hat ;  and 
in  buying  his  goods  by  auction,  as  the  merchant 
often  does,  he  has  not  time  to  make  the  calculation 
before  the  goods  are  cried  off.  He  therefore  lose«i 
the  chance  of  making  good  bargains  by  being1 
afraid  to  bid  at  random,  or  if  he  bids,  and  the 
goods  are  cried  off,  he  may  have  made  a  poor  bar- 
gain, by  bidding  thus  at  a  venture.  It  then  be- 
comes a  useful  and  practical  problem  to  determine 
instantly  what  per  cent,  he  would  gain  if  he  re- 
tailed the  hats  at  a  certain  price. 

To  tell  what  an  article  should  retail  for  t* 
make  a  profit  of  20  per  cent., 

RULE. — Divide  what  the  articles  cost  per  dozen  by 
10,  which  is  done  by  removing  the  decimal  point  onf 
place  to  the  left. 


M  CTLTIPLICTATION.  41 

For  instance,  if  hats  cost  $17.50  per  dozen,  re- 
move the  decimal  point  one  place  to  the  left,  mak- 
ing $1.75,  what  they  should  be  sold  for  apiece  to 
gain  20  per  cent,  on  the  cost.  If  they  cost  $31.00 
per  dozen,  they  should  be  sold  at  $3.10  apiece,  etc. 
We  take  20  per  cent,  as  the  basis  for  the  following 
reasons,  viz. :  because  we  can  determine  instantly, 
by  simply  removing  the  decimal  point,  without 
changing  a  figure;  and,  if  the  goods  would  not 
bring  at  least  20  per  cent,  profit  in  the  home  mar- 
ket, the  merchant  could  not  afford  to  purchase  and 
would  look  for  goods  at  lower  figures. 

REASON. — The  reason  for  the  above  rule  is  ob- 
vious :  For  if  we  divide  the  cost  of  a  dozen  by  12, 
we  have  the  cost  of  a  single  article ;  then  if  we 
wish  to  make  20  per  cent,  on  the  cost,  (cost  being 
1  or  iO>  we  a(^  ^e  ^O  per  cent.,  which  is  £,  to 
the  ^,  making  -|  or  i-J- ;  then  as  we  multiply  the 
cost,  divided  by  12,  by  the  J-f  to  find  at  what  price 
one  must  be  sold  to  gain  20  per  cent.,  it  is  evident 
that  the  12s  will  cancel,  and  leave  the  cost  of  a 
dozen  to  be  divided  by  10,  which  is  done  by  re- 
moving the  decimal  point  one  place  to  the  left. 

1.  If  I  buy  2  doz.  caps  at  $7.50  per  doz.,  wliat 
shall  I  retail  them  at  to  make  20%?     Ans.  75  sts. 

2.  When  a  merchant  retails  a  vest  at  $4.50  and 
makes  20%,  what  did  he  pay  per  doz.?    Ans.  $45. 

3.  At  what  price  should  I  retail  a  pair  cf  boots 
that  cost  $85  per  doz.,  to  make  20%?    Ans.  $8.50. 


42         ORTON'S  LIGHTNING  CALCULATOR. 

RAPID  PROCESS  OF  MARKING  GOODS  AT  DIFFERENT 
PER  CENTS. 

Now,  as  removing  the  decimal  point  one  place 
to  the  left,  on  the  cost  of  a  dozen  articles,  gives 
the  selling  price  of  a  single  one  with  20  per  cent, 
added  to  the  cost,  and,  as  the  cost  of  any  article 
is  100  per  cent.,  it  is  obvious  that  the  selling  price 
would  be  20  per  cent,  more,  or  120  per  cent.; 
hence,  to  find  50  per  cent,  profit,  which  would 
make  the  selling  price  150  per  cent.,  we  would 
first  find  120  per  cent.,  then  add  30  per  cent.,  by 
increasing  it  one-fourth  itself;  to  make  40  per 
cent.,  add  20  per  cent.,  by  increasing  it  one-sixth 
Hself ;  for  35  per  cent.,  increase  it  one-eighth  itself, 
etc.  Hence,  to  mark  an  article  at  any  per  cent, 
profit,  we  have  the  following 

GENERAL  RULE 

First  find  20  per  cent,  profit,  by  removing  the 
decimal  point  one  place  to  the  left  on  the  price  the 
articles  cost  a  dozen;  then,  as  20  per  cent,  profit  is 
120  per  cent.,  add  to  or  subtract  from  this  amount 
the  fractional  part  that  the  required  per  cent,  added 
to  100  is  more  or  less  than  120. 

Merchants,  in  marking  goods,  generally  take  a 
per  cent,  that  is  an  aliqot  part  of  100,  as  25%. 
33^%,  50%,  etc.  The  reason  they  do  this  is  be- 
cause it  makes  it  much  easier  to  add  such  a  per 
cent,  to  the  cost;  for  instance,  a  merchant  could 


MULTIPLICATION.  43 

mark  almost  a  dozen  articles  at  50  per  cent,  profit 
in  the  time  it  would  take  him  to  mark  a  single 
one  at  49  per  cent.  For  the  benefit  of  the  student, 
and  for  the  convenience  of  business  men  in  mark- 
ing goods,  we  have  arranged  the  following  table : 

TABLE 

For  Marking  all  Articles  bought  ~by  the  Dozen. 
N.  B.  Most  of  these  are  used  in  business. 
To  make  20$  remove  the  point  one  place  to  the  left, 

"      "      80$       "         "      «      and  add  J  itself. 

"      "      60$       "          "       "        "      "     4      " 

"      "      50%       "          "      "        "      "     4-      " 

"  "          44:%  "  "  "  "          U         -i-  U 

u      n      40%       a          "       u        "      ((     -i       u 

u        ci 


32*  '   A 

"     "      "    "  A     " 
"     "      "     "  TV     " 

«  a        a          a       u      i          u 

"  "    "  "  X  " 

"      "     subtract^      " 

U  U  U  U  1  a 

w 
If  I  buy  1  doz.  shirts  for  $28.00,  what  shall  I 

retail  them  for  to  make  50%?  Ans.  $3.50. 

EXPLANATION. — Remove  the  point  one  place  to 
the  left,  and  add  on  |  itself. 


44         ORTON'S  LIGHTNING  CALCULATOR. 

Where  the  Multiplier  is  an  Aliquot  Part  of  100. 

Merchants  in  selling  goods  generally  make  the 
price  of  an  article  some  aliquot  part  of  100,  as  in 
Belling  sugar  at  12^  cents  a  pound  or  8  pounds 
for  1  dollar,  or  in  selling  calico  for  16|  cents  a 
yard  or  6  yards  for  1  dollar,  etc.  And  to  be- 
come familiar  with  all  the  aliquot  parts  of  100,  so 
that  you  can  apply  them  readily  when  occasion 
requires,  is  perhaps  the  most  useful,  and,  at  the 
same  time,  one  of  the  easiest  arrived  at  of  all  the 
computations  the  accountant  must  perform  in  the 
practical  calculations  of  the  counting-room. 

TABLE  OF  THE  ALIQUOT  PARTS  OF  100  AND  1000 
N.  B.  Most  of  these  are  used  in  business. 

Ill  is  i-  part  of  100.  8J  is  ^  part  of    100 

25    is  f  or  J  of  100.  16|  is  T2^  or  J-  of    100 

371  ig  I  part  of  100.  33*  is  ^  or  \  of    100 

50    is  f  or  J  of  100.  66|  is  •&  or  f  of    100 

62J  is  f  part  of  100.  83J  is  ^  or  f  of    100. 

75    is  f  or  |  of  100.          125    is  £  part  of  1000, 

87£  is  |  part  of  100.          250    is  f  or  J  of  1000. 

6i  is  Tir  Part  of  10°-  375  is  I  Parfc  of  !000. 
18|  is  T3F  part  of  100.  625  is  f  part  of  1000. 
31i  is  ^  part  Of  100.  875  is  f  part  of  1000. 

To  multiply  by  an  aliquot  part  of  100, 
RULE.- — Add  two  ciphers  to  the  multiplicand,  Jie* 

Mite  such  part  of  it  as  the  multiplier  is  part  of  100, 
N.  B.  If  the  multiplicand  is  a  mixed  number  reduce 

the  fraction  to  a  decimal  of  two  places  before  diTiding. 


MULTIPLICATION.  45 

3.  To  multiply  any  number  by  125  ad  if  three 
ciphers,  and  divide  by  8. 

Multiply  3467  by  125.          Product,  433375. 
8)3467000 

433375 

NOTE. — By  annexing  three  ciphers  the  number 
is  increased  one  thousand  times ;  and  by  dividing 
by  8,  the  quotient  will  be  only  one-eighth  of  1000, 
that  is  125  times. 

4.  To  multiply  any  number  by  16f  add  two 
ciphers,  and  divide  by  6. 

Multiply  3768  by  16f .  Product,  62800. 

6)376800 

62800 

5.  To  multiply  any  number  by  166$  add  three 
ciphers,  and  divide  by  6. 

Multiply  7875  by  166f .      Product,  1312500. 
6)7875000 

1312500 

6.  To  multiply  any  number  by  33J  add  tw* 
ciphers,  and  divide  by  3. 

Multiply  9879  by  33 J.          Product,  329300 
)987900 

329300 


16         ORTON'S  LIGHTNING  CALCULATOR. 

RATIONALE. — As  in  the  last  case,  by  annexing 
two  ciphers,  we  increase  the  multiplicand  one  hun- 
dred times ;  and  by  dividing  the  number  by  3,  we 
only  increase  the  multiplicand  thirty-three  and 
one-third  times,  because  33 J  is  one-third  of  100. 

7.  To  multiply  any  number  by  333J  add  three 
ciphers,  and  divide  by  3. 

Multiply  4797  by  333J.      Product,  1599000. 

3)4797000 

1599000 

8.  To  multiply  any  number  by  6f  add  two  ci- 
phers, and  divide  by  15 ;  or  add  one  cipher  and 
multiply  by  f . 

Multiply  1566  by  6f . 

15)156600 

10440  First  method. 

15660 
2 

3)31320 

10440  Second  method. 

9.  Tc  multiply  any  number  by  66§  add  three 
ciphers,  and  divide  by  15 ;  or  add  two  ciphers  and 
multiply  by  f. 


MULTIPLICATION.  47 


Multiply  3663  by  66f . 

15)3663000 

244200  First  method. 

366300 
2 

3)732600 

244200  Second  method. 

10.  To  multiply  any  number  by  8J  add  two  ci- 
phers, and  divide  by  12. 

Multiply  2889  by  8J.     Product,  24075. 
12)288900 

24075 

11.  To  multiply  any  number  by  83 J  add  thref 
ciphers,  and  divide  by  12. 

Multiply  7695  by  83£.    Product,  641250, 
12)7695000 

641250 

12.  To  multiply  any  number  by  6J  add  two  A 
jhers,  and  divide  by  16  or  its  factors — 4X4. 

Multiply  7696  by  6J.     Product,  4810C 
4)769600 

4)192400 
48100 


48        ORTON'S  LIGHTNING  CALCULATOR. 

13.  To    multiply    any    number    by   62J    add 
three  ciphers,  and  divide  by  16  or  its  factors— 
4x4. 

Multiply  3264  by  62J.    Product,  204000. 
16)3264000 

204000 

14.  To  multiply  any  number  by  18f,  add  two 
ciphers,  and  multiply  by  3,  and  divide  by  16  or  *ts 
factors — 4X4. 

Multiply  768400  by  18f .     Product,  144075. 

768400 
3 


16)2305200 
144075 

15.  To  multiply  any  number  by  31J  add  three 
ciphers,  and  divide  by  32  or  its  factors — 4x8. 

Multiply  7847  by  31J.    Product,  245218g. 
4)7847000 

8)1961750 
245218| 

16.  To  multiply  any  number  by  37  J,  aid  two 
ciphers  and  multiply  by  3,  and  divide  by  8. 


MULTIPLICATION.  49 

Multiply  2976  by  37J.    Product,  111600. 

297600 
3 

8^892800 


111600 

17.  To  multiply  any  number  by  87J  add  two 
ciphers,  divide  by  8,  and  subtract  the  quotient. 
Multiply  6768  by  87J. 

8)676800 
84600 


592200  Ans. 

18.  To  multiply  any  number  by  75  add  two  ci- 
phers, divide  by  4,  and  subtract  the  quotient. 
Multiply  4968  by  75. 

4)496800 
124200 


372600  Ans. 

19.  To  multiply  by   99  add  two  ciphers,  and 
subtract  the  given  number  from  the  result. 
Multiply  31416  by  99. 

3141600 
31416 


Product,  3110184 
As  99  times  is  1  time  less  than  100  times,  and 


60         ORTON'S  LIGHTNING  CALCULATOR. 

adding  00  is  in  effect  multiplying  by  100,  one  time 
the  multiplicand  is  deducted,  which  leaves  99  times, 
as  required. 

This  principle  is  applicable  where  any  number 
of  9's  is  the  multiplier ;  as  many  ciphers  being 
added  of  course  as  there  are  9's.  If  the  multi- 
plier were  98,  we  would  subtract  twice  the  multi- 
plicand ;  if  97,  three  times,  and  so  on. 

20.  To  multiply  by  any  number  of  9's. 

E.ULE. — Annex  as  many  ciphers  to  the  multipli- 
cand as  there  are  9's  in  the  multiplier,  and  from  this 
number  subtract  the  number  to  be  multiplied,  and  th* 
remainder  is  the  product  required. 

NOTE. — To  multiply  by  any  number  of  3's,  pro- 
ceed as  above,  and  divide  the  product  by  3 ;  but  if 
it  be  required  to  multiply  by  6's,  proceed  as  above, 
and  then  multiply  the  product  by  two,  and  divide 
the  result  by  3,  and  the  quotient  is  the  product. 

21.  To  multiply  by  any  number  between  10  and 
20,  as  16  or  18,  multiply  by  the  units'  figure,  and 
set  the  product  under  the  multiplicand ;  but  put  it 
one  place  to  the  right ;  then  add  the  lines  together. 
The  reason  is  evident  on  looking  at  the  calculation. 

Multiply  3854  by  16. 
3854 
23124 


61664  An*. 


MULTIPLICATION.  51 

On  the  same  principle,  if  any  number  of  ct- 
^flers  intervene,  as  106,  1006,  10006,  etc.,  so*  the 
product  so  many  places  farther  to  the  right. 
Multiply  3854  by  1006. 

3854 

23124 


3877124  An*. 

Cross  Multiplication  is  a  mode  of  multiplying 
by  large  multipliers  in  a  single  line ;  and  by  prac- 
tice the  operation  may  be  performed  with  great 
expedition.  It  is  necessary  to  begin  with  small 
numbers,  say  of  two  places,  and  carry  the  calf  dili- 
gently, if  you  would  carry  the  ox  successfully, 

Here  we  multiply  5x2  and  set  down  and  32 
carry  as  usual ;  then  to  what  you  carry  UCkl  45 

5X3  and  4X2,  which  gives  24;  set  down 

4  and  carry  2  to  4x3,  which  gives  14.     It     1440 
is  obvious  that  this  is  just  the  usual  mode, 
with  the  intermediate  work  done   in   the 
head. 

Here  the  first  and  second  places  are  123 
found  as  before;  for  the  third,  add  6X1,  456 

4X3,  5X2,  with  the  2  you  had  to  carry,  

making  30  ;  set  down  0  and  carry  3 ;  then     56088 
drop   the  units'  place   and   multiply  the 
hundreds  and  tens  crosswise,  as  you  did  the  tens 
and  units,  and  you  find  the  thousands'  figure }  then, 
4  E 


62         ORTON'S  LIGHTNING  CALCULATOR. 

dropping  both  units  and  tens,  multiply  the 
adding  the  1  you  carried,  and  you  have  5,  which 
completes  the  product.  The  same  principle  may 
be  extended  to  any  number  of  places ;  but  let  each 
fttep  be  made  perfectly  familiar  before  advancing 
to  another.  Begin  with  two  places,  then  take  three, 
then  four,  but  always  practicing  some  time  on  each 
Bumber,  for  any  hesitation  as  you  progress  will  con- 
fuse you. 

To  Multiply  by  21,  31,  etc.,  to  91  in  a  single 
line,  multiply  by  the  tens'  figure  and  set  the  pro- 
duct one  place  to  the  left  underneath  the  multipli- 
cand ;  then  add. 

Multiply  3854  by  21. 

3854 

7708 

80934  Ans. 

il  ciphers  intervene,  as  201,  3001,  etc.,  multiply 
us  before,  but  set  the  product  as  many  additional 
places  to  the  left  as  there  are  ciphers. 
Multiply  3854  by  6001. 

*  3854 
23134 

2316254  Ans. 
The  following  is  a  convenient  mode  of  multi- 


MULTIPLICATION-.  53 

plying  by  any  two  figures,  and  is  not  difficult  to 
apply: 

Multiply  3754 
By  27 

Product,  101358 

I  here  multiply  27  by  4,  setting  down  the  first 
product  figure  and  carrying  the  others ;  I  then 
multiply  by  5  and  set  down  and  carry  in  the  same 
way — so  proceeding  to  the  highest  place  of  the 
multiplicand. 

Where  the  multiplier  is  not  too  large,  and  can 
be  divided  into  two  or  more  factors,  there  is  a  sav- 
ing in  adopting  the  following  mode,  and  it  may  be 
used  as  a  proof  of  the  common  mode.  If  I  seek 
to  multiply,  say  7864  by  24,  it  requires  in  the 
usual  way  two  product  lines,  and  finding  their  sum 
by  addition,  making  a  third  operation;  but  if  I 
multiply  by  6,  and  that  product  by  4,  or  by  8  and 
3,  or  12  and  2,  the  business  is  dispatched  in  two 
lines.  But  being  able  to  multiply  in  a  single  line 
is  still  better. 

The  same  remarks  apply  in  Division,  and  hence 
there  is  often  economy  of  figures  in  dividing  by 
factors  of  your  divisor ;  and  if  a  remainder  occurs 
only  in  your  first  division  it  is  the  true  one ;  but 
if  in  the  second  only,  then  multiply  such  remain- 
der by  the  first  divisor,  or  all  if  more  than  one ; 


fft         ORTON'S  LIGHTNING  CALCULATOR. 

'*nd  if  there  was  a  remainder  on  the  first  division 
also,  it  must  be  added  in,  and  the  sum  will  be  the 
true  remainder. 

Divide  73640  by  24. 

6 ) 73640 


4)12273  and  2  over. 

3068  and  1  over.  1x6+2=8,  the  true  rem'der. 
Or—     4)73640 

6)18410 

3068  and  2  over,  and  2  x4=8,  as  before. 

Another  mode  of  proving  Division  is  to  divide 
the  dividend  by  the  quotient,  and  the  result  will  be 
the  divisor — the  same  remainder  occurring  in  one 
case  as  in  the  other,  but  not  of  the  same  fractional 
value ;  for  as  the  dividend  exceeds  some  multiple 
of  the  divisor  and  quotient,  just  the  amount  of  the 
remainder,  that  remainder  will  be  the  same,  with- 
out regard  to  which  of  the  factors  occupies  the  di- 
visor's place ;  and  as  the  divisors  would  differ,  th0 
value  of  the  fraction  formed  by  the  remainder  and 
divisor  must  differ.  To  make  the  dividend  an  ex 
tct  multiple,  subtract  the  remainder  from  it. 


MULTIPLICATION.  55 

Multiply  3756 
By    182 

7512 
30048 
3756 

683592 

In  tlie  above  example  the  second  product  Lne 
may  be  conveniently  found  by  multiplying  the  first 
by  4 ;  but  if  there  is  an  error  in  the  first  it  will  run 
into  the  second.  A  shorter  mode  would  be  to  mul- 
tiply the  first  product  line  by  9,  which  would  give 
the  product  by  18  in  one  line,  and  save  work.  Trj 
it. 

As  a  large  number  of  the  problems  solved  ii, 
this  book  are  solved  by  different  modes,  they  fur- 
rdsh  a  great  variety  of  modes  of  proof. 

To  multiply  a  decimal  or  a  mixed  number  by  10, 
100,  1000,  etc.,  remove  the  decimal  point  one,  two, 
or  three  places  to  the  right ;  and  to  divide  by  such 
numbers  remove  the  point  as  many  places  to  the 
left ;  and  if  there  be  not  a  sufficient  number  of 
places  on  the  left,  ciphers  must  be  prefixed. 
.  To  multiply  or  divide  by  any  composite  number ; 
multiply,  or  divide,  as  the  case  may  be,  by  the  fac- 
tors of  the  numbers  successively. 

This  was  sufficiently  illustrated  under  the  head 
of  Proofs. 


56         ORION'S  LIGHTNING  CALCULATOR. 

Tlie  French  mode  of  operating  in  Long  Dh  ision 
has  some  advantage  over  ours.  They  place  the 
divisor  on  the  right  of  the  dividend,  as  we  do  the 
quotient,  and  place  the  quotient  underneath  the 
divisor,  by  which  the  figures  to  be  multiplied 
together  are  brought  near  each  other.  Thus : 

Divid.     Divis. 

3936)96 

384    41  Quotient. 

96 
96 

For  sake  of  brevity  they  fre- 
qtiently  omit  the  product  figures, 
setting  down  only  the  remain-         3936)96 
ders,  which  they  find   as  they  96  41,  Ans. 

pass  along.     Thus :  — 

This,  however,  applies  to  our  mode  as  well  as 
theirs. 

The  following  mode  of  multiplying"^  large 
multipliers  in  a  single  line,  is  both  curious  and 
useful : 

Multiply  7865  by  432  in  a  single  line. 

On  a  slip  of  paper,  separate  from  that  on  which 
the  multiplicand  is  written,  place  the  multiplier  in 
Inverted  order :  thus,  234  and  close  to  the  upper 
edge  of  the  paper.  Then  bring  the  multiplier  so 
that  the  2  shall  be  directly  under  the  5,  or  units' 


MULTIPLICATION.  57 

place  of  the  multiplicand:  multiply  those  figures, 
set  down  0  and  carry  1.  Slide  the  paper  to  the 
left  one  place,  that  2  may  be  under  G,  and  3  under 
5  ;  and  to  the  1  you  carried  add  the  products  of 
the  2  by  6,  and  3  by  5,  making  28 — set  down  8 
and  carry  2.  Again  move  your  paper  one  place  to 
the  left,  and  to  the  2  you  carried,  add  the  several 
products  of  the  multiplicand  figures  with  the 
figures  of  the  multiplier  that  are  under  them,  viz: 
8X2,  6X3,  5X4,  and  the  result  will  be  56 ;  set 
down  6  and  carry  5.  Slide  again  and  you  have  5 
(that  you  carried)  +14+24+24=67.  Thus  pro- 
ceed toward  the  left  until  the  multiplier  passes  from 
under  the  multiplicand,  each  time  adding  what  you 
carry,  to  the  several  products  of  the  figures  that 
stand  one  over  the  other,  the  result  will  be  3397680. 
These  additions  will  soon  be  performed  at  a  glance, 
as  the  products  are  obvious  without  the  formality 
of  naming  the  factors. 

To  understand  these  directions  clearly,  factors 
must  be  placed  on  slips  of  paper,  and  the  directions 
strictly  complied  with ;  by  which  the  mode  of  op- 
eration and  the  reason  will  be  better  understood  in 
ten  minutes,  than  in  three  hours  without  them. 
When  familiar  with  the  slide,  the  operator  may 
proceed  without  it,  and  perform  operations  aston- 
ishing to  tke  uninitiated;  the  largest  numbers 
being  multiplied  together  readily  in  a  single  liae. 


68         ORION'S  LIGHTNING  CALCULATOR. 


01  THE  PROPERTIES  OF  NUMBERS. 

PROPOSITION  1. 

Digits  in  our  system  of  notation  increase  in 
oaluefrom  right  to  left  in  a  tenfold  ratio. 

PROPOSITION  2. 

rn  any  series  of  digits  expressing  a  number,  the 
value  of  any  digit  is  greater  than  the  value  of  all 
the  digits  on  its  right. 

This  property  results  also  from  value  according 
lo  place ;  and  that  the  proposition  is  true  is  obvi- 
mis,  for  if  we  take  the  smallest  digit  (1)  and  place 
it  on  the  left  of  the  largest  (9)  we  form  19;  the  1 
expresses  10  units,  while  the  9  expresses  but  9 
units;  and  let  us  add  what  numbers  of  nine  we 
may,  the  unit  will  constantly  retain  its  greater 
value:  e.g.  19,  199,  1999,  etc.  Not  only  is  the 
left  hand  digit  higher  in  value  than  all  upon  its 
right,  but  the  same  remark  applies  to  each  digit, 
in  reference  to  those  on  its  right. 
PROPOSITION  8. 

If  the  sum  of  the  digits  in  any  number  be  a  multi- 
ple of  9,  the  whole  number  is  a  multiple  of  9. 
This  is  one  of  several  peculiar  properties  of  the 
number  9,  all  arising  from  its  being  just  one  less 
than   the  radix   of  our  system  of  notation,  and 


MULTIPLICATION.  59 

hence  the  highest  number  expressed  by  a  single ' 
character;  and  these  properties  will  belong  to  the 
highest  number  so  expressed  in  any  system.  We 
might  go  a  step  further  in  reference  to  this  prop- 
erty, and  say  that  it  belongs  to  any  number  that 
will  divide  the  radix  of  the  system,  and  leave  one 
as  a  remainder. 

If  we  carefully  examine  the  genesis  of  numbers, 
we  must  see  that  so  far  as  the  number  9  is  con- 
cerned, this  is  an  accidental  property,  resulting 
from  our  scale  of  notation.  We  constantly  express 
each  successive  number  from  unity  to  9,  inclusive, 
by  a  digit  of  greater  value  than  any  preceeding  it ; 
but  when  we  pass  9  we  express  the  next  number, 
10,  by  a  unit  and  a  cipher.  The  number  is  one 
greater  than  9,  and  the  sum  of  its  digits  is  1. 
Eleven  is  2  greater,  and  the  sum  of  its  digits  is 
1-j-l— 2.  Thus  we  proceed,  the  sum  of  the  dig- 
its constantly  expressing  the  excess  over  9,  until 
we  reach  18,  or  twice  9.  Nineteen  is  1  and  9,  and 
it  is  one  over  twice  9.  20  is  2  over  twice  9,  and 
the  sum  of  its  digits  is  2.  The  same  course  con 
tinued  to  millions,  would  but  produce  the  same 
recurring  result.  Nine  is  1  less  than  10;  twice 
9  are  2  less  than  20;  3X9  are  3  less  than  30  ,  and 
BO  on;  and  hence  the  1  of  10,  2  of  20,  3  of  30 
etc.,  come  just  in  the  proper  place  to  keep  up  the 
excess  above  9  and  its  multiples.  If  tr  e  multiples 


60         ORION'S  LIGHTNING  CALCULATOR. 

of  9  did  not  constantly  fall  at  each  product,  one 
further  behind  the  corresponding  multiples  of  10, 
th*  two  of  20,  3  of  30,  etc.,  would  not  fall  in  the 
right  place,  to'  keep  up  the  regular  order  of  the 
aeries. 

PROPOSITION  4. 

If  tlie  sum  of  the  digits  in  any  number  be  a  multi- 
ple of  3,  the  number  is  a  multiple  of  3. 
The  same  reasoning  applied  to  the  number  9  to 
*how  the  correctness  of  the  preceding  proposition, 
will  show  the  correctness  of  this.     Ten,  the  sum 
of  whose  digits  is  1,  is  1  over  3  times  3;  11,  the 
sum  of  whose  digit  is  2,  is  2  more  than  3  times  3 ; 
12,  the  sum  of  whose  digit  is  3,  is  a  multiple, 
etc.,  etc. 

PROPOSITION  5. 

Dividing  any  number  ty  9  or  3,  will  leave  the  same 
remainder  as  dividing  the  sum  of  its  digits  by  9 
or  3. 

This  proposition  follows  as  a  matter  of  course 
from  the  two  next  preceding  it ;  and  we  shall  ad- 
duce .no  other  proof  of  its  correctness.  Like  the 
former,  it  is  an  accidental  property  of  the  highest 
number  expressed  by  a  single  digit  in  any  system, 
and  of  all  its  factors.  If  9  were  the  basis  of  our 
system,  these  properties  would  belong  to  8,  4,  and 
2;  if  8,  then  7  only,  since  7  has  no  factors;  and 


MULTIPLICATION.  6 1 

if  7  were  the  basis,  then  6,  3,  and  2  would  possess 
these  properties;  and  if  12  were  the  basis,  then  11 
only  would  possess  such  properties;  for  it  would 
in  that  case  be  expressed  by  a  single  digit,  and 
would  be  the  highest  number  so  expressed.  Twelve 
would  be  written  with  a  unit  and  a  cipher  as  10 
now  is ;  and  11  being  prime,  it  would  be  the  only 
number  that  would  divide  the  radix  of  the  system 
and  leave  1  as  a  remainder. 

As  early  as  1657,  Dr.  Wallis,  of  England, 
applied  this  principle  to  prove  the  correctness  of 
operations  in  the  elementary  rules  of  Arithmetic, 
and  the  practice  has  been  continued  to  the  present 
time.  The  operation  is  performed  thus : 

We  cast  the  nines  out  of  each    Add  79864=7 
number  separately,  and  set  the  ex-  32075=8 

cess  on  the  right.     We  then  cast  83214=0 

the  nines  out  of  the  sum  total  61840=1 

305160,  and  also  out  of  the  sum  48167=8 

of  the  excesses  8+1+8+7,  and 
they  are  equal :  both  being  6,  and  305160=6 

we  hence "  infer  that  the  work  is 
right.  To  cast  out  the  nines,  the  number  may  be 
divided  by  9 ;  but  a  better  way  is  to  add  the  digits 
together  in  each  number,  rejecting  9  whenever  it 
occurs,  and  carrying  forward  only  the  excess.  Thus 
7  and  8  are  15;  9  being  rejected,  we  carry  6  to  d 
is  12;  rejecting  9,  we  carry  3  to  4=7;  the  num- 


62         ORTON'S  LIGHTNING  CALCULATOR. 

ber  carried  in  each  place  is  the  excess  over  9 ;  ana 
where  9  occurs  it  is  passed  over. 

In  Subtraction  cast  out  the  nines  from  the  minu- 
end and  subtrahend,  and  also  from  the  remainder 
If  the  excess  in  the  remainder  is  equal  to  the  dif- 
ference of  excesses  in  the  minuend  and  subtrahend, 
the  work  is  right. 

Here,  as  we  can  not  take  8  From  6894321=6 
from  6,  we  take  from  9  and  Take  2960864=8 

add  6;  the  result,  7  agrees 

with  the  excess  above  9  in  Leaves  3933457=7 
the  difference  of  the  numbers. 

In  Multiplication,  find  the  excess  in  the  factors, 
and  if  the  excess  in  the  product  of  these  two  ex- 
cesses equals  the  excess  in  the  product  of  the  fac- 
tors the  operation  is  correct. 

Multiply        48756=3 
By  245=2 


243780  3 

195024  - 

97512  6 


11945220=6 


This  is  often  called  proving  by  the  cross ;  and 
instead  of  placing  the  excessss  after  marks  of 
equality,  they  are  placed  in  the  angles  of  a  cross 
as  on  the  right  hand  of  the  above  operation. 


MULTIPLICATION.  63 

In  Division  cast  the  nines  out  of  the  divisor, 
lividend,  quotient,  and  remainder;  then  to  the 
product  of  the  excesses  in  the  divisor  and  quo- 
tient, add  the  excess  in  the  remainder,  and  cast  the 
nines  out  of  the  sum,  and  if  the  excess  equal  that 
in  the  dividend  the  work  is  right. 

Excess  in  Divisor  0 

Excess  in  Quotient  8 

27)465  — 

Product  of  Excess  0 

17+6       Add  Excess  of  Remainder  6 

Excess  in  Dividend  6=6 


Hence  the  work  is  right,  the  excesses  being  equal. 
It  is  proper  to  remark  that  this  mode  of  proof 
is  liable  to  much  objection.  If  the  figures  become 
transposed,  or  if  mistakes  are  made  that  balance 
each  other,  the  work  will  prove  right  when  it  is 
wrong.  The  work  will,  however,  never  prove 
wrong  when  it  is  right.  In  the  product  above, 
you  may  transpose  the  digits  as  you  please,  the 
work  will  prove,  since  the  excess  is  the  same  what- 
ever is  the  order  of  the  digits ;  and  ciphers  may  al- 
ways be  omitted.  Or  if  mistakes  balance  each 
other,  as  if  instead  of  91145  it  be  83216.  The 
excess  here  will  be  the  same  and  the  work  will 
prove,  though  not  a  figure  is  right. 


64         ORION'S  LIGHTNING  CALCULATOR. 

PROPOSITION  6. 

If  from  any  number  the  sum  of  its    digits  le  sub* 
tractedj  the  remainder  is  a  multiple  of  9. 

Fm  31416  For, 

Take      15=3+1+4+1+6  31416^-9=3490+6 

15+9=       1+6 

9)31401  

31401-7-9=3489 


3489  times  9,  diff. 


As  the  remainder  on  dividing  the  given  number 
by  9  will  be  just  the  same  as  on  dividing  the  sum 
of  its  digits,  (Prop.  5,)  it  is  obvious  that  the  differ- 
ence must  be  an  even  multiple.  The  given  num- 
ber is  6  more  than  3490  times  9 ;  the  sum  of  digits 
is  6  more  than  1  time,  hence  their  difference  is 
3489  times. 

This  principle  is  sometimes  used  as  a  puzzle; 
you  may  let  a  person  write  down  any  num-     6745 
ber  for  a  minuend,   then  have  the   party        22 

add  the  figures  and  place  their  sum  for  a    

subtrahend;  then,  after  subtracting,  let  the  6723 
person  rub  out  any  one  figure  in  the  re- 
mainder and  give  you  the  figures  left  in  the  re- 
mainder. You,  by  adding  the  figures  given  and 
subtracting  their  sum  from  the  next  multiple  of  9, 
may  tell  the  figure  rubbed  out,  although  you  did 
not  see  a  figure  that  was  written.  Try  it. 


MULTIPLICATION.  65 

PROPOSITION  7. 

The  difference  between  a  given  number  and  the  dig- 
its composing  such  number  reversed  or  any  how  . 
arranged,  is  always  a  multiple  of  9. 
The  difference,  for  instance,  between  7425,  and 
any  arrangemsut  you  can  make  of  the  same  figures 
is  a  multiple  of  9. 

From     7425  7425  7425 

Take  '   5247  5724  2457 


9)2178          9)1701          9)4968 
242  189  552 


This  is  based  on  the  same  reason  as  the  preced- 
ing; for  whether  you  take  the  sum  of  the  digits  or 
transpose  the  digits,  it  is  the  same  in  effect.  The 
excess  over  an  even  multiple  being  the  same  as 
in  the  given  number,  the  difference  must  nec- 
essarily be  an  even  multiple. 

A  practical  application  is  sometimes  made  of  this 
principle  by  a  person  setting  down  two  rows  of 
Ggures  for  subtraction,  but  being  careful  to  have 
the  figures  of  the  subatrahend  and  minuend  the 
same,  though  differently  arranged.  One  figure  of 
the  remainder  is  then  stricken  out,  and  the  puzzlo 
is  to  restore  it  without  seeing  the  minuend  and  sub- 
trahend It  is  done  by  taking  such  number  as  will 
make  the  remainder  a  multiple  of  9. 


66  ORTON^S   LIGHTNING   CALCULATOR. 

Here  if  5,  4,  or  1,  be         From    7351681 
erased,  any  one  may  restore         Take      1864537 

it ;  but  if  the  9  or  0  be  re-  

moved   he   can    not   know  5490144 

whether   a   9    or  a   cipher  

should  be  supplied,  as  either  will  make  the  num> 
ber  a  multiple  of  9. 

The  mode  we  have  adopted  in  explaining  the 
four  last  preceding  propositions  appears  to  us  plain 
and  sufficiently  satisfactory. 

In  addition  to  the  use  of  these  properties  as 
modes  of  proof,  they  are  the  key  to  many  numeral 
puzzles  and  amusing  questions ;  and  hence  the  care 
we  have  bestowed  in  explaining  the  principle. 
What  has  been  said  may  be  a  sufficient  explanation 
of  the  following  article  on  the  "  Wonderful  Proper- 
ties  of  the  Number  Nine :  " 

"  Multiply  9  by  itself  or  any  other  digit,  and  the 
figures  of  the  product  added  will  be  9. 

Take  the  sum  of  our  numerals  1+2+3+4+5+ 
6+7+8+9=45,  the  digits  of  which,  4+5=9. 
Multiply  each  of  these  digits  by  9,  and  their  sum 
will  be  405  ;  which  added  4+0+5=9 ;  and  405-7- 
9=45,  also  a  multiple  of  9. 

Multiply  any  number,  large  or  small,  by  9,  or 
9  times  any  digit,  and  the  sum  of  the  digits  i)f  the 
product  will  be  a  multiple  of  9. 

Multiply  the  9  digits  in  their  order,  123456 


MULTIPLICATION.  67 

7  6  9,  by  9,  or  any  multiple  of  9  not  exceeding  9 
times  9,  and  the  product,  except  the  tens'  place, 
will  be  all  the  same  figures,  while  the  tens'  place 
will  be  filled  with  0.  Th;3  significant  figure  will 
always  be  the  number  of  times  9  is  contained  in 
the  multiplier. 

27,  or  3  times  9,  will  123456789 

produce  all  3s;  4  times  9  18=9X2 

all  4s. 

Omit  8  in  the  multi-  987554312 

plicand  and  the  product         123456789 

will  be  all  the  same  dig-         

its,  the  0  having  disap-         2222222202 
peared." 

To  a  superficial  observer  the  above  results  may 
Beem  accidental,  but  investigation  will  show  that 
they  all  flow  from  the  laws  and  principles  we  have 
laid  down ;  and  that  a  much  longer  list  might  be 
made  of  apparently  simple  and  detached  facts,  but 
really  of  results  flowing  from  well  established  laws. 
There  are  no  unaccountable  properties  in  numbers. 

While  the  number  9  has  some  peculiar  proper- 
ties from  being  the  next  below  the  radix  of  the 
system,  the  number  11  has  some  peculiarities  from 
being  next  above  the  radix.  Among  these  are  the 
following : 

"  If  from  any  number  the  sum  of  tho  digits 

standing  in  the  odd  places  be  subtracted,  and  to 
5 


68         ORTON'S  LIGHTNING  CALCULATOR. 

the  remainder  the  sum  of  the  digits  standing  in 
the  even  places  be  added,  then  the  result  is  a  mul- 
tiple of  11."  Again,  "If  the  sum  of  the  digits 
standing  in  the  even  places  be  equal  to  the  sum  of 
the  digits  standing  in  the  odd  places,  or  differ  by 
11  or  any  of  its  multiples,  the  number  is  a  multiple 
of  11." 

As  these  however  are  of  no  practical  utility,  we 
shall  not  discuss  them. 

The  number  7  has  also  some  peculiarities,  but 
we  shall  name  only  one,  as  they  are  useless.  If  a 
number  be  divided  into  periods  of  three  figures  each, 
beginning  at  the  units1  place,  when  the  difference  of 
the  sums  of  the  alternate  periods  is  a  multiple  of  7, 
the  whole  number  is  a  multiple  of  7. 

Here  862—428  is  a  mul-     7)382,907,428,862 

tiple  of  7 ;  and  so  is  907 — 

382;   therefore   the  whole  54,701,061,266 

number  is  a  multiple  of  7. 

The  division  of  numbers  into  Even  and  Odd 
seems  to  arise  from  considering  them  in  pairs. 
The  following  facts  growing  out  of  this  division 
will  be  readily  understood  : 

The  sum  of  two  even  numbers  is  even,  and  so  ia 
their  difference  :  8+4=12;  8 — 4=4. 

The  sum  of  an  odd  number  of  odd  numbers  is 
odd;  but  the  sum  of  an  even  number  of  odd  num- 
bers is  even :  3+5+7=15;  3+5=8;  and  5+7=1 2. 


MULTIPLICATION.  69 

An  even  and  an  odd  number  being  added  to- 
gether, or  one  subtracted  from  the  other,  the  result 
will  be  odd:  8+3—11;  8—3=5. 

If  a  number  has  0,  2,  4,  6  or  8  in  the  units' 
place,  it  is  divisible  by  2,  and  is  consequently  even. 

No  odd  number  can  be  divided  by  an  even  num- 
ber without  a  remainder. 

If  an  odd  number  measure  an  even  one,  it  will 
also  measure  the  half  of  it.  7  measures  42,  and 
therefore  measures  21,  the  half  of  it. 

If  the  sum  of  the  digits  standing  in  the  EVEN 
places,  be  equal  to  the  sum  of  the  digits  standing  in 
the  ODD  places,  then  the  number  is  divisible  by  II. 

Thus  the  numbers  121,  363,  12133,  48422,  etc., 
are  all  divisible  by  11. 

Multiplication  and  Division. 

.  To  multiply  one -half,  is  to  take  the  multiplicand 
one-half  of  one  time;  that  is,  take  one-half  of  it, 
or  divide  it  by  2. 

To  multiply  by  J,  take  a  third  of  the  multipli- 
cand, that  is,  divide  it  by  3, 

To  multiply  by  f ,  take  J,  first,  and  multiply  that 
by  2;  or,  multiply  by  2  first,  and  divide  the  pro- 
duct by  3.* 

*Sometimes  one  operation  is  preferable,  and  sometimes 
the  other;  good  judgment  alone  can  decide  when  thi 
case  is  bef>re  us. 


70          ORTON'S  LIGHTING  CALCULATOR. 

EXAMPLES. 

I.  What  will  360  barrels  of  flour  come  to  at  5J 
dollars  a  barrel.  At  1  dollar  a  barrel  it  would  be  360 
dollars ;  at  5 J  dollars,  it  would  be  5J  times  as  much. 
360 


5  times,  1800 

J  of  a  time,        90 


Am.  $1890 

Before  we  attempt  to  divide  by  a  mixed  number, 
such  as  2J,  3J,  5f  ,  etc.,  we  must  explain,  or  rather 
observe  the  principle  of  division,  namely:  That 
the  quotient  will  be  the  same  if  we  multiply  the  divi- 
dend and  divisor  by  the  same  number.  Thus  24 
divided  by  8,  gives  three  for  a  quotient.  Now,  if 
we  double  24  and  8,  or  multiply  them  by  any  num- 
ber wnatever,  and  then  divide,  we  shall  still  have  3 
for  a  quotient.  16)48(3;  32)96(3,  etc. 

Now,  suppose  we  have  22  to  be  divided  by  5J; 
we  may  double  both  these  numbers,  and  thus  be 
clear  of  the  fraction,  and  have  the  same  quotient. 
5|)22(4  is  the  same  as  11)44(4. 

How  many  times  is  1J  contained  in  12?  Ans. 
Just  as  many  times  as  5  is  contained  in  48.  The 
5  is  4  times  1  J,  and  48  is  4  times  12.  From  these 
observations,  we  draw  the  following  rule  for  divid- 
ing by  a  mixed  number. 


MULTIPLICATION   AND   DIVISION.  71 

RULE. — Multiply  the  whole  number  "by  the  lower 
term  of  the  fraction ;  add  the  upper  term  to  the  pro- 
duct  for  a  divisor ;  then  multiply  the  dividend  by 
the  lower  term  of  tJie  fraction,  and  then  divide. 

How  many  times  is  1^  contained  in  36?  An*. 
30  times. 

N.  B.  If  we  multiply  both  these  numbers  by  5, 
they  will  have  the  same  relation  as  before,  and  a 
quotient  is  nothing  but  a  relation  between  two 
numbers.  After  multiplication,  the  numbers  may 
be  considered  as  having  the  denomination  of  fifths. 

How  many  times  is  J  contained  in  12  ?  Ans.  48 
times. 

One-fourth  multiplied  by  4,  gives  1;  12,  multi- 
plied by  4,  gives  £8.  Now,  1  in  48  is  contained  48 
times. 

Divide  132  by  2|.  Ans.  48. 

Divide  121  by  15  J.  Ans.    8 

How  many  times  is  f  contained  in  3  ?  Ans.  4 
times. 

By  a  little  attention  to  the  relation  of  numbers, 
we  may  often  contract  operations  in  multiplication. 
A  dead  uniformity  of  operation  in  all  cases  indi- 
cates a  mechanical  and  not  a  scientific  knowledge 
of  numbers.  As  a  uniform  principle,  it  is  much 
easier  to  multiply  by  the  small  numbers,  2,  3,  4, 
5,  than  by  7,  8,  9. 


72         ORTON'S  LIGHTNING  CALCULATOR. 

Multiply    4532        Multiply    4532 
by  639  by  963 

(63=9X7.)        40788  40788 

285516  285516 


Product,  2895948  Product,  4364316 

In  both  the  foregoing  examples  we  multiply  the 
product  of  9  by  7,  because  7  times  9  are  equal  to  63. 
Because  9  is  in  the  place  of  hundreds  in  exam- 
ple 2,  the  product  for  the  other  two  figures  is  set 
two  places  toward  the  right. 

In  this  last  example  we  may  commence  with  the 
3  units  in  the  usual  way ;  then  that  product  by  2, 
because  2  times  3  are  6  ;  then  the  product  of  3  by 
3,  which,  will  give  the  same  as  the  multiplicand  by 
9.  The  appearance  of  the  work  would  then  be  the 
same  as  by  the  usual  method,  but  would  be  easier, 
as  we  actually  multiply  by  smaller  numbers. 

Multiply    40788 
by  497 


285516 
1998612 


20271636 


Product  of  the  7  units. 
As  7X7=49,  multiply  the 
product  of  7  by  7. 


Every  fact  of  this  kind,  though  extremely  sim- 
ple, should  be  known  bj  all  who  seek  for  knowl- 
edge in  figures. 


MULTIPLICATION   AND   DIVISION. 


73 


First  multiply  by  12, 
then  that  product  by  12. 


Multiply  785460 
by          14412 

9425520 
113106240 


11320049520 

Multiply       576  Multiply    this     last 

by  186         number,  3456,  (which 

is  6  times  576,)  by  3> 

(6X3     18.)  3456         and  place  the  product 

10368  in   the   place   of  tens, 

and  we  have  180  times 

107136  576.  Observe  the  same 
principle  in  the  follow- 
ing examples : 

Multiply        576          Multiply    4078£ 
by  618  by 


Commence  with  6. 
(6X3=18.) 


497 


3456   (7X7=49.)  285516 
10368         1998612 


355968 


20271636 


Multiply  61524 
by     7209 


553716 
4429728 

Product,  443646516 


Multiply  this  pro- 
duct of  9  by  8,  because 
9  times  8  are  72,  and 
place  the  product  in 
the  place  of  100,  be* 
cause  it  is  7200. 


74         ORTON'S  LIGHTNING  CALCULATOR. 

Multiply      1243 
by  636 

7458  First  by  600. 

44748        Multiply  7458  by  6. 

Product,        790548 

Multiply     7  8  6  '4 
by  246 

This  may  be  done  by  commencing  with  the  2; 
tnen  that  product  by  2  and  3 ;  or  we  may  com- 
mence with  the  6  units,  and  then  that  product  by 
4 ;  because  4  times  6  are  24. 
Multiply  3764  by  199. 

Take  3764  200  times,  and  from  that  product  sub- 
tract 3764. 

Multiply  764  by  498J. 

Take  764  500  times,  and  from  that  product  sub- 
tract 1|  times  764. 

Multiply  396  by  21|,  or,  (which  is  the  same,) 
99X87=8700  —86     8613. 

N.  B.— Ninety-nine  is  J  of  396,  and  87  is  4 
times  2 If. 

How  many  times  is  125  contained  in  2125  ? 
Same  as  250  in  4250 ; 
Same  as  25  in  425 . 
Same  as  50  in  850  , 
Same  as  5  in  85  ; 
Same  as  10  m  170  j  that  is,  17  timeu. 


MULTIPLICATION  AND   DIVISION.  75 

The  object  of  these  changes  is  to  give  the 
learner  an  accurate  and  complete  knowledge  of 
numbers  and  of  division;  and  the  result  is  not  the 
only  object  sought  for,  as  many  young  learners 
suppose. 

How  many  times  is  75  contained  in  575  ?  or  di- 
dde  575  by  75.  An*.  7f . 

Divide  800  by  12J.  Quotient,  64. 

Divide  27  by  16f .  Quo.  3  T%,  or  If  J. 

A  person  spent  6  dollars  for  oranges,  at  6J  cents 
a-piece;  how  many  did  he  purchase?  Ans.  96. 

"When  two  or  more  numbers  are  to  be  multiplied 
together,  and  one  or  more  of  them  having  a  cipher 
on  the  right,  as  24  by  20,  we  may  take  the  cipher 
from  one  number  and  annex  it  to  the  other  with- 
out affecting  the  product ;  thus,  24x20  is  the  same 
as  240X2;  286X1300—28600X13;  and  350X 
70x40=35x7x4X1000,  etc. 

Every  fact  of  this  kind,  though  extremely  simple, 
will  be  very  useful  to  those  who  wish  to  be  skillful 
in  operation. 

NOTE. — If  there  are  ciphers  at  the  right  hand 
either  of  the  multiplier  or  multiplicand,  or  of  both, 
they  may  be  neglected  to  the  close  of  the  opera- 
tion, when  they  must  be  annexed  to  the  product. 

REMARKS.  —We  now  give  a  few  examples,  for  the  pur- 
pose of  teaching  the  pupil  how  to  use  his  judgment;  he 
will  then  have  learned  a  rule  more  valuable  than  aU  others. 
G 


76         ORTON'S  LIGHTNING  CALCULATOR. 


Multiplication  and  Division   Combined. 

WHEN  it  becomes  necessary  to  multiply  two  or 
more  numbers  together,  and  divide  by  a  third,  or 
by  a  product  of  a  third  and  fourth,  it  must  be  lit* 
erally  done  if  the  numbers  are  prime. 

For  example :  Multiply  19  by  13  and  divide  that 
product  by  7. 

This  must  be  done  at  full  length,  because  the 
numbers  are  prime  ;  and  in  all  such  cases  there  will 
result  a  fraction. 

But  in  actual  business  the  problems  are  almost  all 
reduceable  by  short  operations ;  as  the  prices  of 
articles,  or  amount  called  for,  always  corresponds 
with  some  aliquot  part  of  our  scale  of  computation. 
And  when  two  or  more  of  the  numbers  are  composite 
numbers,  the  work  can  always  be  contracted. 

Example :  Multiply  375  by  7,  and  divide  that 
product  by  21.  To  obtain  the  answer,  it  is  suffi- 
cient to  divide  375  by  3,  which  gives  125. 

The  7  divides  the  21,  and  the  factor  3  remains 
for  a  divisor.  Here  it  becomes  necessary  to  lay 
down  &  plan  of  operation. 

Draw  a  perpendicular  Hue  and  place  all  numbers 
that  are  to  be  multiplied  together  under  each  other, 
on  the  right  hand  side,  and  all  numbers  that  are 
Uvisors  under  each  other,  on  the  left  hand  side. 


MULTIPLICATION   AND   DIVISION.  77 

EXAMPLES. 

Multiply  140  by  36,  and  divide  that  product  by 
84.  We  place  the  numbers  thus  : 

84  I  14° 
4  I    36 

We  may  east  out  equal  factors  from  each  side  of 
the  line  without  affecting  the  result.  In  this  case 
12  will  divide  84  and  36  j  then  the  numbers  will 
Btand  thus : 

140 

3 

But  7  divides  140,  and  gives  20,  which,  multi- 
plied by  3,  gives  60  for  the  result. 

Multiply  4783  by  39,  and  divide  that  product 
by  13. 

r*    4783 
*0  3 

Three  times  4783  must  be  the  result. 
Multiply  80  by  9,  that  product  by  21,  and  di- 
vide the  whole  by  the  product  of  60 X  6X14. 


In  the  above  divide  60  and  80  by  20,  and  14  and 
21  by  7,  and  those  numbers  will  stand  canceled  as 
above,  with  3  and  4,  2  and  3,  at  their  sides. 

Now,  the  product  3X@X2,  on  the  divisor  side, 
is  equal  to  4  times  9  on  the  other,  and  the  remain' 
ing  3  is  the  result. 


78         ORTON'S  LIGHTNING  CALCULATOR. 


General  Rules  for  Cancellation. 

RULE  IST.  Draw  a  perpendicular  line;  obsetre 
this  line  represents  the  sign  of  equality.  On  the 
right  hand  side  of  this  line  place  dividends  only; 
on  the  left  hand  side  place  divisors  only ;  having 
placed  dividends  on  the  right  and  divisors  on  the 
left,  as  above  directed. 

2d.  Notice  whether  there  are  ciphers  both  on 
the  right  and  left  of  the  line ;  if  so,  erase  an  equal 
number  from  each  side. 

3d.  Notice  whether  the  same  number  stands 
both  on  the  right  and  left  of  the  line ;  if  so,  erase 
them  both. 

4th.  Notice  again  if  any  number  on  either  side 
of  the  line  will  divide  any  number  on  the  opposite 
side  without  a  remainder ;  if  so,  divide  and  erase 
the  two  numbers,  retaining  the  quotient  figure  on 
the  side  of  the  larger  number. 

5th.  See  if  any  two  numbers,  one  on  each  side, 
can  be  divided  by  any  assumed  number  without  a 
remainder;  if  so,  divide  them  by  that  number,  and 
retain  only  their  quotients.  Proceed  in  the  same 
manner,  as  far  as  practicable,  then, 

6th.  Multiply  all  the  numbers  remaining  on  the 
rignt  hand  side  of  the  line  for  a  dividend,  am* 
those  remaining  on  the  left  for  a  divisor. 

7th    Divide,  and  the  quotient  is  the  answer. 


INTEREST,    DISCOUNT   AND   AVERAGE.          79 

NOTE. — If  only  one  number  remain  on  either 
iide  of  the  line,  that  number  is  the  dividend  or 
divisor,  according  as  it  stands  on  the  right  or  left 
of  the  line  The  figure  1  is  net  regarded  in  the 
operation,  because  it  avails  nothing,  either  to  mul- 
tiply or  divide  by. 

REMARKS. — This  method  may  not  work  a  great 
many  problems,  as  they  are  found  in  some  books, 
but  it  will  work  90  out  of  every  100  that  ought  to 
be  found  in  books. 

In  a  book  we  might  find  a  problem  like  this : 

What  is  the  cost  of  21b.  7oz.  13pwt.  of  tea,  at 
7s.  5d.  per  pound.  But  the  person  who  should  go 
to  a  store  and  call  for  31b.  7oz.  and  13pwt.  of  tea 
would  be  a  fit  subject  for  a  mad-house.  The  above 
problem  requires  downright  drudgery,  which  every 
one  ought  to  be  able  to  perform ;  but  such  drudgery 
never  occurs  in  business. 


INTEREST,  DISCOUNT,  AND  AVERAGE. 

BEFORE  entering  upon  an  investigation  of  the 
different  modes  of  calculating  interest,  it  may  be 
interesting  to  bestow  some  attention  upon  the  his- 
tory of  the  subject,  that  we  may  be  better  prepared 
to  understand  it. 


80         ORTON'S  LIJHTNING  CALCULATOR. 

Among  the  Jews  a  law  existed  that  they  should 
not  lake  interest  of  their  brethren,  though  they 
wero  permitted  to  take  it  of  foreigner*.  "  Thou 
sh?Jt  not  lend  upon  usury  to  thy  brother:  usury 
of  money,  usury  of  victuals,  usury  of  any  thing 
that  is  lent  upon  usury ;  unto  a  stranger  thou  may- 
cst  ]end  upon  usury ;  but  unto  thy  brother  thou 
ehalt  not  ^nd  upon  usury."  (Deuteronomy  xxiii, 
19,  20.)  Alter  the  dispersion  of  the  Jews  they 
wandered  through  the  earth,  but  they  yet  remain 
a  distinct  people,  mixing,  but  not  becoming  assim- 
ilated with  the  people  among  whom  they  reside. 
Still  looking  to  the  period  when  they  shall  return 
to  the  promised  land,  they  seldom  engage  in  per- 
manent business,  but  pursue  traffic,  and  especially 
dealing  in  money ;  and  if  their  national  policy  for- 
bids their  taking  interest  of  each  other,  they  show 
no  backwardness  in  taking  it  unsparingly  of  the 
rest  of  mankind.  For  ages  they  have  been  the 
money  lenders  of  Europe,  and  we  may  safely  at- 
tribute to  this  circumstance  the  prejudice,  in  some 
measure,  that  still  exists  even  in  our  own  country 
against  such  as  pursue  this  business  as  a  profession. 
The  prejudice  of  the  Christian  against  the  Jew  has 
been  transferred  to  his  occupation,  and  from  the 
days  of  Shakspeare,  who  painted  the  inexorable 
Shylock  contending  for  his  pound  of  flesh-,  down 
Jx>  the  present  time,  the  grasping  money  lender,  no 


INTEREST,    DISCOUNT,    AND   AVERAGE.          81 

less  than  the  grinding  dealer  in  other  matters,  has 
been  sneeringly  called  a^ew. 

For  ages  the  taking  of  any  corapensatioii  wnLt- 
ever  for  the  use  of  money  was  called  usury,  and 
was  denounced  as  unchristian  ;  and  we  find  Aris- 
totle, the  heathen  philosopher,  gravely  contending 
that  as  money  could  not  beget  money,  it  was  bar- 
ren, and  usury  should  not  be  charged  for  its  use. 
The  philosopher  forgot  that  with  money  the  bor- 
rower could  add  to  his  flocks  and  his  fields,  and 
profit  by  the  produce  of  both. 

Definition  of  Terms. 

Interest  is  premium  paid  for  the  use  of  money, 
{roods,  or  property. 

It  is  computed  by  per  centage — a  certain  per 
cent.  OB  the  money  being  paid  for  its  use  for  a 
stated  tirce.  The  money  on  which  interest  is  paid 
is  called  the  PRINCIPAL. 

The  per  cent,  paid  is  called  the  RATE  ;  the  prin- 
cipal and  interest  aaded  together  is  called  the 
AMOUNT. 

When  a  rate  per  cent,  is  stated,  without  tho 
mention  of  any  term  of  time,  tn^  time  is  under- 
v  stood  to  be  1  year. ' 

The  first  important  step  in  the  calculation  of 
simple  interest  is  the  arranging  of  tne  tm:e  for 
which  it  is  computed.  The  student  must  study  the 


82  ORTON^S    LIGHTNING   CALCULATOR. 

following  Propositions  carefully,  if  lie  would  be 
expert  in  this  important  and  useful  branch  of  bus- 
Sness  calculations : 

PROPOSITION  1. 

/  the  time  consists  of  years,  multiply  the  principal 
ly  the  rate  per  cent.,  and  that  product  by  the 
number  of  years. 

EXAMPLE  1. — Find  the  interest  of  $75  for  4 
tears  at  6  per  cent. 
Operation. 

$75  The  decimal  for  6  per  cent,  is 

.06  06.     There  being  two  places  of 

decimals  in  the  multiplier,  we 
4.50  point  off  two  in  the  product. 

4 

818.00  4ns. 

PROPOSITION  2. 
If  the  time  consists  of  years  and  months,  reduce  the 

time  to  months,  and  multiply  the  principal  by  the 

rate  per  cent,  and  number  of  months  together,  and 

divide  the  result  by  12. 

NOTE. — The  work  can  always  be  abbreviated  at 
I,  6,  8,  9,  12,  and  15  per  cent.,  by  canceling  the 
per  cent.,  or  time,  or  principal,  with  the  common 
divisor  12. 


INTEREST,    DISCOUNT,    ^ND 


S3 


EXAMPLE  2.— Find  the  interest  of  $240  for  2 
years  and  7  months  at  8  per  cent. 
First  method.  Second  method  : 


Principal,             $240 
Per  cent.,                 .08 

In.  forlyr.,         19.20- 
2yrs.-j-7nios.,           Slmos. 

by  cancellation. 
^0—20 
It       8            rate. 
31           time. 

49.60  Ans. 

12)595.20 
$49.60  Ans. 

The  operation  by  canceling  is  much  more  brief. 
We  simply  place  the  principal,  rate,  and  time,  on 
the  right  of  the  line,  and  12  on  the  left;  then  we 
cancel  12  in  240,  and  the  quotient  20  multiplied 
with  8  and  31  gives  the  interest  at  once. 

NOTE. — After  12  is  canceled  the  product  of  the 
remaining  numbers  is  always  the  interest. 

PROPOSITION  3. 

If  the  time  consists  of  years,  months,  and  days,  re- 
duce the  years  to  months,  add  in  the  given  months, 
and  place  one-third  of  the  days  to  the  right  of 
this  number,  which  we  multiply  by  the  principal 
and  rate  per  cent.,  and  divide  by  12,  as  before; 
or  cancel  and  divide  by  12  before  multiplying. 
EXAMPLE  3. — Find  the  interest  of  $231  for  1 

year,  1  month,  and  6  days,  at  5  per  cent. 


84 


ORTON'S  LIGHTNING  CALCULATOR, 


First  method. 
Principal, 
Per  cent., 

In.  for  lyr., 


$231 
.05 

11.55 
,    13.2mo, 


12)152.460 


Second  method : 
by  cancellation. 
231          prin. 
5  rate. 

— 11 


$12.705  Ans. 


$12.705  Am. 

By  the  second  method  we  cancel  12  in  132,  and 
multiply  the  quotient  11  by  5  and  231. 

NOTE. — When  the  principal  is  $,  and  the  time  is 
in  years  or  months,  the  interest  is  in  cents ;  if  the 
time  is  in  years,  months,  and  days,  the  interest  is 
in  mills,  unless  the  days  are  less  than  3,  in  which 
case  it  would  be  in  cents,  as  before. 

NOTE. — The  reason  we  divide  the  days  by  3  is 
because  we  calculate  30  days  for  a  month,  and  di- 
viding by  3  reduces  the  days  to  the  tenth  of  months, 

NOTE. — The  three  preceding  propositions  will 
work  any  note  in  interest  for  any  time  and  at  any 
given  rate  per  cent. 

How  to  Avoid  Fractions  in  Interest. 

PROPOSITION  4. 

If,  when  the  time  consists  of  years,  months,  and 
days,  are  not  divisible  by  3,  you  can  divide  the  days 
ly  3,  and  annex  the  mixed  number  as  in  Proposition 


INTEREST,  DISCOUNT,  AND  AVERAGE. 


3,  or  if  you  wish  to  avoid  fractions,  you  can  reduce 
the  time  to  interest  days,  and**tnultiply  the  principal^ 
rate  and  days  together,  and  divide  the  result  by  36 
or  its  factors,  4x9. 

NOTE. — In  this  case  as  in  the  preceding,  the 
work  can  almost  always  be  contracted  by  dividing 
the  rate  or  time  or  principal  with  the  divisor  36. 

NOTE. — We  use  the  divisor  36,  because  we  cal- 
culate 360  interest  days  to  the  year.  We  discard 
the  0,  because  it  avails  nothing  to  multiply  or  di- 
vide by. 

EXAMPLE  4.— Find  the  interest  of  $210  for  1 
year,  4  months,  and  8  days,  at  9  per  cent. 

Year.        Months.        Days. 

1  4          8=16.2f  months  or  488  days. 

Operation  Operation 

By  Prop.  3.  By  Prop.  4. 

$210  $210 

.9 


18.90 
16.2J 


12)307440 
$25.620  Ans. 


18.90 

488 


36)922320 
$25.620  Ans. 


We  will  now  work  the  example  by  cancellation 
to  show  its  brevity. 


66          ORTON'S  LIGHTING  CALCULATOR. 

Operation  ty  Cancellation. 
Time  488  days. 

210 

0 

m  122 

122 

210 


$25.620 

£*W  cancel  9  in  36  goes  4  times,  then  4  into  485 
122.  Now  multiply  remaining  numbers  to- 
gether, thus,  210X122  and  we  have  the  interest  at 
once. 

When  the  days  are  not  divisible  by  3  we  reduce 
t\i3  whole  time  to  days ;  then  we  place  the  princi- 
pal rate  and  time  on  the  right  of  the  line.  Now, 
because  the  time  is  in  days,  we  place  36,  on  the 
left  of  the  line  for  a  divisor.  (If  the  time  was 
months  we  would  place  12  on  the  left.) 

NOTE. — A  very  short  method  of  reducing  time 
to  interest  days  is  to  multiply  the  years  by  36 ; 
add  in  3  times  the  number  of  months  and  the  tens' 
figure  of  the  days,  and  annex  the  unit  figure;  but 
if  the  days  are  less  than  10  simply  annex  them. 

EXAMPLE  1. — Reduce  1  year,  2  months,  and  6 
days,  to  days. 

Years.        Months.  Days. 

36X1+3X2=:42  annex  6=426  Ant. 


SIMPLE  INTEREST  BY  CANCELLATION.          87 

EXAMPLE  2. — Reduce  2^ears,  3  months    and 
17  days,  to  interest  days. 

Years.    M'ths.  Days.  .    Days. 

36x2+3x3+1=82.  annex  7=827  days  An*. 
NOTE. — The  student  should  commit  to  memory 
the  multiplication  of  the  number  36  up  as  far  as 
9  times   36,  and  then  he  can  reduce  almost  in- 
stantly years,  months,  and  days,  to  days. 


SIMPLE  INTEREST  BY  CANCELLATION, 

RULE. — Place  the  principal,  time,  and  rate  per 
cent,  on  the  right  hand  side  of  the  line.  If  the  time 
consists  of  years  and  months,  reduce  them  to  months, 
and  place  12  (the  number  of  months  in  a  year)  on 
the  left  hand  side  of  the  line.  Should  the  time  con- 
sist of  months  and  days,  reduce  them  to  days  or  deci- 
mal parts  of  a  month.  If  reduced  to  days,  place 
36  on  the  left.  If  to  decimals  parts  of  a  month, 
place  12  only  as  before. 

Point  off  two  decimal  places  when  the  time  is  in 
months,  and  three  decimal  places  when  the  time  u 
in  days. 

NOTE.  If  the  principal  contains  cents,  point  off 
four  decimal  places  when  the  time  is  in  months, 

and  five  decimal  places  when  the  time  is  in  days. 
H 


88         ORTON'S  LIGHTNING  CALCULATOR. 

NOTE. —  We  place  36  on  the  left  because  there  u 
360  interest  days  in  a  year.  (Custom,  has  made  thu 
lawful.} 

EXAMPLE  1. — What  is  the  interest  on  $60  for 
117  days  at  6  per  cent? 
Operation. 


Here  117X0 
must  be  the  $ 
answer. 


Both  sixes    on    the 
right  cancels  36  on 


117  the    left,    and    we 

— —  have    nothing    left 

$1.170  Ans.     to  divide  by. 
In  this  case  we  point  off  three  decimal  places  be- 
cause the  time  is  in  days.    If  the  time  had  been  117 
months,  we  would  have  pointed  off  but  two  deci- 
mal places. 

EXAMPLE  2.— What  is  the  interest  of  $96.50 
for  90  days  at  6  per  cent? 
Operation. 

96.50  9650 

00 — 15  15 

A  ___ 

1.44.750  Ans. 

Now  cancel  6  in  36  and  the  quotient  6  into 
90,  and  we  have  no  divisor  left.  Hence  15X96.50 
must  be  the  answer. 

NOTE — As  there  are  cents  in  the  principal,  we 
point  off  five  decimals  ;  three  for  days  and  two  for 
cents.  Pay  no  attention  to  the  decimal  poot  Until 
the  clo  of  the  operation. 


SIMPLE  INTEREST  BY  CANCELLATION.    S9 

EXAMPLE  3. — What   is   the   interest   of  $480 
ftr  361  days  at  6  per  cent? 

**0— 80  361 

361  80 


$28.880  Ans. 
Now  cancel  6  in  36  and  the  quotient  6  into  480, 
and  we  have  no  divisor  left.     Hence  80X361  must 
be  the  answer. 

EXAMPLE  4.-— What  is  the  interest  of  $720  for 
9  months  at  7  per  cent? 

W0— 60  60 

9  9 

7  _ 

540 

7 


$37.80  Ans. 

Now  cancel  12  in  720  there  is  nothing  left  to 
divide  by.  Hence  60X9X7  must  be  the  answer. 

N.  B.  When  interest  is  required  on  any  sum  for 
days  only,  it  is  a  universal  custom  to  consider  30 
days  a  month,  and  12  months  a  year ;  and,  as  the 
unit  of  time  is  a  year,  the  interest  of  any  sum  for 
one  day  is  3^0,  what  it  would  be  for  a  year.  For 
2  days,  3 1 0,  etc.;  hence  if  we  multiply  by  the 
days,  we  must  divide  by  360,  or  divide  by  36  and 
save  labor.  The  old  form  of  this  method  was  to 
place  360,  or  12  and  30,  on  the  left  of  the  line, 
but  using  36  is  much  shorter. 


90  ORTONS   LIGHTNING   CALCULATOR. 

WHEN   THE   DAYS   ARE  NOT  DIVISIBLE   BY  THREE 

NOTE. — When  the  time  consists  of  months  and 
days,  and  the  days  are  not  divisible  by  three,  re- 
d\ce  the  time  to  days. 

EXAMPLE  5. — What  is  the  interest  of  $960  foi 
11  months  and  20  days  at  6  per  cent? 

Months.    Days. 

Operation.  11     20=350  days. 

000—160  350 

0  —36     350  160 

$56.000 

Now  cancel  6  in  36  and  the  quotient  6  into 
960,  and  we  have  no  divisor  left.  Hence  160X 
350  must  be  the  answer. 

EXAMPLE  6. — What  is  the  interest  of  $173  for 
8  months  and  16  days  at  9  per  cent? 

Months.    Days. 

Operation.  8     16=256  days. 

173  173 

0  64 


$11.072  Ans. 

Now  cancel  9  in  36  and  the  quotient  4  into  256, 
and  we  have  no  divisor  left.  Hence  64X173  must 
"be  the  answer. 

N.  B.  Let  the  pujil  remember  that  this  is  a  gen- 
eral and  universal  method,  equally  applicable  to 
any  per  cent,  or  any  required  time,  and  all  ri  'ier 
rules  must  be  reconcilable  to  it;  and,  in  ft"*  *1I 
other  rules  are  but  modifications  of  this. 


SIMPLE   INTEREST   BY    CANCELLATION.         91 


EXAMPLE  7. — What  is  Jhe  interest  on  $1080 
A)r  7  months  and  11  days  at  7  per  cent? 

Months.    Days. 

7         11=221  days. 

Operation. 

0#0—  30  221 

221  30 

7  _, 

6630 
7 


$46,410  Ans. 

Now  cancel  36  in  1080  and  we  have  no  divisor 
left,  hence  30X221X7  must  be  the  answer. 

WITH  MORE  DIFFICULT  TIME  AND  RATE  PER  CENT. 

EXAMPLE  8.  —  What  is  the  interest  of  $160  for 
19  months  and  23  days  at4J  per  cent? 

Months.    Days. 

19         23=593  days. 
Operation. 

160—20  593 

593  20 


$11.860  Ans. 
Now  cancel  4J  in  36  and  the  quotient  8  into  160 
we  have  no  divisor  left,  hence1  20x5  93  must  be 
the  interest. 

WHEN    THE   DAYS   ARE    DIVISIBLE    BY    THREE. 

EULE.  —  Place  one-third  of  the  days  to  the  right 
v/  the  months,  and  place  12  on  the  left  of  the  line. 


92         ORTON'S  LIGHTNINQ  CALCULATOR. 

NOTE. — Dividing  the  days  by  3  reduces  them  to 
tenths  of  months,  as  3  days— 1  tenth  of  a  month, 
6  days=2  tenths,  etc. 

EXAMPLE  9. — What  is  the  interest  of  $240  for 
1  year  11  months  and  12  days  at  5  per  cent? 

Years.    Months.    Days. 

1  11       12=23.4  months. 
Operation. 

&I0—20  20 

234  5 

5  

100 
234 

$23.400  Am. 

Now  Cancel  12  in  240  and  we  have  no  divisor 
left,  hence  20x5X234  must  be  the  interest. 

EXAMPLE  10. — What  is  the  interest  of  $500  for 
2  years  5  months  and  24  days  at  6  per  cent  ? 

Yeare.    Months.    Days. 

2  5        24=z29.8  months. 
Operation. 

500  149 

20$— 149  500 

$74.500  Ans. 

Now  cancel  6  in  12  and  the  quotient  2  into  298 
and  we  have  no  divisor  left,  hence  500x1  *9  equals 
the  interest. 


SIMPLE  INTEREST  Bl  CANCELLATION.    93 

EXAMPLE  11. — What  is  the  interest  of  $350  for 
3  years  7  months  and  6  days  at  10  per  cent? 

Years.    Months.    Days. 

3         7          6=43.2  months. 

Operation. 

350  350 

It    4#2— 36  36 

10  

12600 
10 

$126.000  Ans. 

Now  cancel  12  in  432  and  we  have  no  divisor 
left.  Hence  350x36X10  equals  the  interest. 

EXAMPLE  12.— What  is  the  interest  of  $241  for 
13  months  and  9  days  at  8  per  cent? 

Months.    Days. 

13         9=13.  3  months. 
Operation. 

241  241 

13.3  133 


32053 
2 


3)64106 

$21.368f  Ans. 

In  this  example  I  canceled  8  and  12  by  4,  and 
then  multiplied  all  on  the  right  of  the  line  and  di 


P4         ORTON'S  LIGHTNING  CALCULATOR. 

vided  by  3.  If  I  could  have  divided  by  3  before 
multiplying  I  would  have  saved  labor,  but  when  the 
numbers  are  prime  the  whole  work  must  be  liter* 
ally  done. 

CLOSING  REMARKS. — We  have  now  fully  ex- 
plained the  canceling  system  of  computing  inter- 
est. Any  and  every  problem  can  be  stated  by  this 
method,  and  the  beauty  and  simplicity  of  the 
system  ranks  it  high  among  the  most  important 
abbreviations  ever  discovered  by  man.  As  we  have 
before  remarked,  at  6,  4,  8,  9,  12,  15,  and  4J  per 
cents.,  every  problem  in  interest  can  be  canceled, 
besides  a  great  many  can  be  abbreviated  at  5,  7,  and 
other  per  cents.;  and  after  the  problem  has  beec 
stated  and  we  find  that  we  can  not  cancel,  what 
have  we  done?  We  have  simply  stated  the  prob- 
lem in  its  simplest  and  easiest  form  for  working  it 
by  any  other  method.  Hence  we  have  a  decided 
advantage  of  all  notes  that  will  cancel,  and  if  we 
can  not  cancel  we  have  stated  the  problem  in  its 
correct  and  proper  form  for  going  through  tie 
whole  work ;  but  it  is  only  when  the  principal^ 
time,  and  rate  per  cent,  are  all  prime,  that  the 
WHOLE  work  must  be  LITERALLY  done.  At  6  per 
sent,  we  can  cancel  through,  and  6  is  the  rate  most 
commonly  used 


6HORT  PRACTICAL  RULES.         95 


SHORV    PRACTICAL    RULES, 

DEDUCED  FROM  THE  CANCELING  SYSTEM, 

For  calculating  interest  at  6  per  cent.,  either  for 
months,  or  months  and  days. 

To  find  the  interest  for  months  at  6  per  cent. 

RULE. — Multiply  the  principal  by  half  the  num- 
ber of  months,  expressed  decimally  as  a  per  cent.; 
that  is,  for  12  months,  multiply  by  .06  ;  for  8  months, 
multiply  by  .04. 

NOTE  1. — It  is  obvious  that  if  the  rate  per  cent, 
were  12,  it  would  be  1  per  cent,  a  month  ;  if,  there- 
fore, it  be  6  per  cent.,  it  will  be  a  half  per  cent,  a 
month  ]  that  is,  half  the  months  will  be  the  per 
cent. 

NOTE  2. — If  any  other  per  cent,  is  wanted  you 
can  proceed  as  above,  and  then  multiply  by  the 
given  rate  per  cent,  and  divide  by  6,  and  the  quo- 
tient is  the  interest. 

1.  Whtt  is  the  interest  of  $363  for  8  inoflths? 

$368 
.04=half  the  months. 


NOTE  3. — When  the  months  are  not  even ;  that  **< 
will  not  divide  by  2,  multiply  oiw  Ix^f  tkc  f  rv*^  2 


96         ORTON'S  LIGHTNING  CALCULATOR. 

by  tl.e  whole  number  of  months,  expressed  deci- 
mally. 

To  find  the  interest  of  any  sum  at  6  per  cent, 
per  annum  for  any  number  of  months  and  days. 

RULE. — Divide  the  days  by  3  and  place  the  quo- 
tient to  the  right  of  the  months;  one-half  of  the  num- 
ber thus  formed  multiplied  by  the  principal,  or  one- 
half  of  the  principal  multiplied  by  this  number,  will 
give  the  interest — pointing  off  three  decimal  places 
when  the  principal  is  $. 

2.  What  is  the  interest  of  $76  for  1  year,  5 
months,  and  12  days,  at  6  per  cent? 

Years.  Months.  Days. 

1        6        12=18.4  months— half  9,2. 
$76  Or,  184 

9.2  38=half  prin. 

$6.992  An*.  $6.992  Ans. 

NOTE. — Dividing  the  days  by  3  reduces  them  to 
the  tenth  of  months. 

To  find  the  the  interest  of  any  sum  at  6  per 
cent,  per  annum  for  any  number  of  days. 

RULE. — Divide  the  principal  by  6  and  multiply 
the  quotient  by  the  number  of  days;  or  divide  the 
days  by  6  and  multiply  the  quotient  by  the  principal, 
pointing  off  three  decimal  places  when  the  principal 
**. 

NOTE. — Always  divide  6  into  the  number  that 


SHORT  PRACTICAL  RULES.         97 

will  divide  without  a  remainder;  if  neither  one 
will  divide,  multiply  the  principal  and  days  to- 
gether and  divide  the  result  by  6. 

3.  What  is  the  interest  of  $240  for  18  days  at 
6  per  cent  ? 

18-.-6—3  240-^-6=40 

$240  Or,  $40=4 of  Prin- 

3z=J  of  the  days.  18 

$0.720  Ans.  $0.720  Ans. 

4.  What  is  the  interest  of  $1800  for  72  days  at 
6  per  cent. 

$1800  Or,  $300=£  of  prin. 

12=%  of  the  days.  72 

$21.600  Ans.  $21.600  Ans. 

Useful  Suggestions  to  the  Accountant  in  Computing 
Interest  at  6  per  cent. 

If  the  principal  is  divisible  by  6,  always  reduce 
the  time  to  days;  then  multiply  the  number  of 
days  by  one-sixth  of  the  principal. 

EXAMPLE. 

5.  Find  the  interest  of  $240  for  1  year,  5  months, 
and  17  days,  at  6  per  cent. 

6)240          lyr,  5mos.,  17da.=527  days. 

Multiplied  by    40 
$  of  prin.~40 

$21.080  An*, 


90         ORTON'S  LIGHTNING  CALCULATOR. 

If  the  days  are  only  divisible  by  3,  multiply 
one -third  of  the  principal  by  one-half  of  the  days. 

6.  What  is  the  interest  of  $210  for  80  days  at  6 
per  cent.  ? 

$70=J  of  the  principal. 
40=|  of  the  days. 

§2.800  An*. 

When  the  Rate  of  Interest  is  4  per  cent 
RULE. — Multiply  the  principal  by  one-third  the 
number  of  months,  or  by  one-ninth  the  number  of 
days,  and  the  product  is  the  interest. 

NOTE. — This  principle  is  also  deduced  from  the 
canceling  method  of  computing  interest ;  the  stu- 
dent can  readily  see  that  4  is  J  of  12  and  ^  of  36, 

When  the  Rate  of  Interest  is  9  per  cent. 
RULE. — Multiply  the  principal  by  three-fourth* 
(he  number  of  months,  or  one-fourth  the  number  of 
days,  or  vice  versa. 


BANKERS    METHOD  CF  COMPUTING  INTEREST.  99 

BANKEBS'   METHOD 

OF 

COMPUTING    INTEREST, 

AT  6  PER  CENT.  FOR  ANY  NUMBER  OF  DAYS. 


RULE. — Draw  a  perpendicular  line,  cutting  off 
the  two  right  hand  figures  of  the  $,  and  you  have  tht 
interest  of  the  sum  for  60  days  at  6  per  cent. 

NOTE. — The  figures  on  the  left  of  the  line  are  $, 
and  those  on  the  right  are  decimals  of  $. 

EXAMPLE  1. — What  is  the  interest  of  $423 
60  days  at  6  per  cent.  ? 

$423=the  principal. 

$4  |  23  cts.=interest  for  60  days. 

NOTE. — When  the  time  is  more  or  less  than  60 
days,  first  get  the  interest  for  60  days,  and  from 
that  to  the  time  required. 

EXAMPLE  2. — What  is  the  interest  of  $124  for 
1  >  -days  at  6  per  cent.  ? 

Days.  Day§. 

15= J  of  60 

$124— principal. 

4)1  |  24  cts.=interest  for  60  days. 

|  31  cts. ^interest  for  15  days. 
7  I 


100          ORTON'S  LIGHTNING  CALCULATOR. 

EXAMPLE  3. — What  is  the  interest  of  $123.40  fo* 
90  days  at  6  per  cent.? 

Days.  Days.  Days, 

90=60+30 
$123.40=principal. 
2)1  I  2340=interest  for  60  days. 
|  6170=interest  for  30  days. 

Ans.   $1  |  851=iinterest  for  90  days. 

EXAMPLE  4. — What  is  the  interest  of  $324  for 
75  days  at  6  per  cent.? 

Days.  Days.  Days. 

$324=rprincipal.  75=60+ 15 


4)3 


24  cts.  interest  for  60  days. 
81  cts.  interest  for  15  days. 


8.  $4  |  05  cts.  interest  for  75  days. 
REMARKS. — This  system  of  Computing  Interest 
is  very  easy  and  simple,  especially  when  the  days 
are  aliquot  parts  of  60,  and  one  simple  division 
will  suffice.  It  is  used  extensively  by  a  large  ma- 
jority of  our  most  prominent  bankers ;  and,  indeed, 
is  taught  by  most  all  Commercial  Colleges  as  the 
snortest  system  of  computing  interest. 

Method  of  Calculating  at  Different  Per  Cents. 

This  principle  is  not  confined  alone  to  6  per  cent, 
as  many  suppose  who  teach  and  use  i\*  It  is  their 
custom  first  to  find  the  interest  at  6  per  cent.,  and 
from  that  to  other  per  cents.  But  it  is  equally  ap- 
plicable for  all  per  cents.,  from  1  to  15  inclusive. 


BANKERS1  METHOD  OF  COMPUTING  INTEREST.  101 

The  following  table  shows  the  different  per  cents., 
with  the  time  that  a  given  number  of  $  will 
amount  to  the  same  number  of  cents  when  placed 
at  interest. 

RULE. — Draw  a  perpendicular  line,  cutting  off 
the  two  right  hand  figures  of  $,  and  you  have  th« 
interest  at  the  following  percents. 

Interest  at  4  per  cent,  for  90  days. 

Interest  at  5  per  cent,  for  72  days. 

Interest  at  6  per  cent,  for  60  days. 

Interest  at  7  per  cent,  for  52  days. 

Interest  at  8  per  cent,  for  45  days. 

Interest  at  9  per  cent,  for  40  days. 

Interest  at  10  per  cent,  for  36  days. 

Interest  at  12  per  cent,  for  30  days. 

Interest  at  7-30  per  cent,  for  50  days. 

Interest  at  5-20  per  cent,  for  70  days. 

Interest  at  10-40  per  cent,  for  35  days. 

Interest  at  7J  per  cent,  for  48  days. 

Interest  at  4J  per  cent,  for  80  days. 
NOTE. — The  figures  on  the  left  of  the  perpen- 
dicular line  are  dollars,  and  on  the  right  decimals 
of  $.     If  the  $  are  less  than  10  prefix  a  0. 

EXAMPLE  1. — What  is  the  interest  of  $120  for 
1 5  days  at  4  per  cent? 

Days,  Bays, 

$120— principal.  15=r£  of  90. 

6)1     20  cts.~  int  for  90  days. 
20  cts.— int.  for  15  days. 


1  D2        ORTON'S  LIGHTNING  CALCULATOR. 

EXAMPLE  2. — What  is  the  interest  of  $132  for 
13  days  at  7  per  cent.  ? 

Days.  Days. 

$132=prmcipal.  13=J  of  52. 

4)1  I  32  cts.=int.  for  52  days. 

I  33  cts.=int.  for  13  days.  9 
EXAMPLE  3. — What  is  the  interest  of  $520  for 
9  days  at  8  per  cent.  ? 

Days.  Days. 

$520=principal.  9=r£  of  45. 

20  cts.r=int.  for  45  days. 
04  cts.r=int.  for  9  days. 

EXAMPLE  4.— What  is  the  interest  of  $462  for 
for  64  days  at  7J  per  cent.  ? 

Days.  Days.  Days. 

$462=principal.  64=48+16. 


3)4 


62  cts.  =int.  for  48  days. 
54  cts.  =iut.  for  16  days. 


$6  |  16  cts.=int.  for  64  days. 

REMARK. — We  have  now  illustrated  several  ex- 
amples by  the  different  per  cents. ;  and  if  the  stu- 
dent will  study  carefully  the  solution  to  the  above 
examples,  he  will  in  a  short  time  be  very  rapid  in 
this  mode  of  computing  interest. 

NOTE. — The  preceding  mode  of  computing  in- 
terest is  derived  and  deduced  from  the  canceling 
system ;  as  the  ingenious  student  will  readily  see. 
It  is  a  short  and  easy  way  of  finding  interest  for 
days  when  the  days  are  even  or  aliquot  parts ;  but 
when  they  are  not  multiples,  and  three  or  four  di 


BANKERS'  METHOD  OP  COMPUTING  INTEREST.  103 

visions  are  ncessary,  the  canceling  system  is  much 
more  simple  and  easy.  We  will  here  illustrate  an 
example  to  show  the  difference :  Eequired  the  in- 
terest of  $420  for  49  days  at  6  per  cent. 

Bankers'  method.  Canceling  meth. 


2)4 

2)2 
5)1 
3) 


0—70 


20  cts.i=int.  for  60  days. 

10  cts.=int.  for  30  days. 
05  cts.— int.  for  15  days. 

21  cts.— int.  for  3  days. 

7  cts.nrint.  for  1  day.  $3.430  Ans. 


49 
70 


$3  |  43  cte.r^int.  for  49  days. 

The  canceling' method  is  much  more  brief;  we 
simply  cancel  6  in  36,  and  the  quotient  6  into  420  ; 
there  is  no  divisor  left;  hence  70x49  gives  the  in- 
terest at  once. 

If  the  time  had  been  15  or  20  days,  the  Bankers' 
Method  would  have  been  equally  as  short,  because 
15  and  20  are  aliquot  parts  of  60.  The  superiority 
the  canceling  system  has  above  all  others  is  this :  it 
takes  advantage  of  ths  principal  as  well  as  the  time. 

For  the  benefit  of  the  student,  and  for  the  con- 
venience of  business  men,  we  will  investigate  this 
system  to  its  full  extent  and  explain  how  to  take 
advantage  of  the  principal  when  no  advantage  can 
be  taken  of  the  days.  This  is  one  of  the  most  im- 
portant characteristics  of  interest,  and  very  often 
eaves  much  labor.  It  should  be  used  when  the  day* 
are  not  even  or  aliquot  parts. 


104          ORTON'S    LIGHTNING   CALCULATOR. 

The  following  table  shows  the  different  sums  of 
money  (at  the  different  per  cents.)  that  bear  1  cent 
interest  a  day ;  hence  the  time  in  days  is  always 
the  interest  in  cents ;  therefore,  to  find  the  interest 
*  on  any  of  the  following  notes  at  the  per  cent,  at- 
tached to  it  in  the  table,  we  have  the  following 
rule: 

RULE. — Draw  a  perpendicular  line,  cutting  off 
the  two  right  hand  figures  of  the  days  for  centsy  and 
you  have  the  interest  for  the  given  time. 

Interest  of  §90  at  4  per  cent,  for  1  day  is  1  cent 

Interest  of  §72  at  5  per  cent,  for  1  day  is  1  cent. 

Interest  of  §60  at  6  per  cent,  for  1  day  is  1  cent. 

Interest  of  $52  at  7  per  cent,  for  1  day  is  1  cent. 

Interest  of  $45  at  8  per  cent,  for  1  day  is  1  cent. 

Interest  of  $40  at  9  per  cent,  for  1  day  is  1  cent. 

Interest  of  $36  at  10  per  cent,  for  1  day  is  1  cent 

Interest  of  $30  at  12  per  cent,  for  1  day  is  1  cent. 

Interest  of  $50  at  7.30  per  ct.  for  1  day  is  1  ct. 

Interest  of  $70  at  5.20  per  ct.  for  1  day  is  1  ct. 

Interest  of  $35  at  10.40  per  ct.  for  T  day  is  1  ct. 

Interest  of  $48  at  7-J  per.  cent,  for  1  day  is  1  cent. 

Interest  of  $80  at  4J  per  cent,  for  1  day  is  1  cent. 

Interest  of  $24  at  15  per  ct.  for  1  day  is  1  cent. 

NOTE. — The  7.30  Government  Bonds  are  calcu- 
lated on  the  base  of  365  days  to  the  year,  and  the 
5.20's  and  10.40's  on  the  base  of  364  days  to  the  year 


BANKERS   METHOD  OF  COMPUTING  INTEREST.    105 

NOTE. — This  table  should  be  committed  to  mem- 
ory, as  it  is  very  useful  when  the  days  are  not  even 
or  aliquot  parts.  If  the  days  are  less  than  10  pre- 
fix a  0  before  drawing  the  line. 

EXAMPLE  1. — Required  the  interest  of  $60  for 
117  days  at  6  per  cent. 

11 7= the  days.  Here  we  cut  off  the  two 

$1  |  17  cts.  Ans.  right  hand  figures  for  cents. 

The  student  should  bear  in  mind  that  the  inter- 
est on  $60  for  117  days  is  just  the  same  as  the 
interest  on  $117  for  60  days. 

By  looking  at  the  table  we  see  that  the  interest 
for  $60  at  6  per  cent,  is  1  cent  a  day ;  hence  the 
time  in  days  is  the  answer  in  cents.  If  this  note 
was  $120,  instead  of  $60,  we  would  first  find  the 
interest  for  $60,  and  then  double  it;  if  it  was 
$180,  we  would  multiply  by  3,  etc. 

EXAMPLE  2. — Required  the  interest  of  $45  for 
219  days  at  8  per  cent. 

219=the  days.  Here  we  cut  off  the  two 

$2  |  19  cts.  Ans.  right  hand  figures  for  cents. 

The  student  should  bear  in  mind  that  the  inter- 
est on  $45  for  219  days  is  just  the  same  as  the 
interest  on  $219  for  45  days. 

By  looking  at  the  table  we  see  that  the  interest 
on  $45  at  8  per  cent,  is  1  cent  a  day ;  hence  the 
time  in  days  is  the  answer  in  cents.  If  this  Dote 


106       ORTON'S  LIGHTNING  CALCULATOR. 

was  $22.50,  instead  of  $45,  we  would  first  get  the 
interest  for  $45,  and  then  divide  by  2 ;  if  it  was 
$75^  we  would  add  on  f ;  if  $60,  add  on  J,  etc. 

EXAMPLE  3.— Required  the  interest  of  $48  for 
115  days  at  9  per  cent. 

115—the  days.  $48=$40+$8. 

5)$1  I  15  cts.=the  int.  of  $40  for  115  days. 
I  23  cts.=the  int.  of  $8  for  115  days. 

Ans.  $1  |  38  cts.^the  int.  of  $48  for  115  days. 

Here  we  first  find  the  interest  of  $40,  because 
the  days  is  the  interest  in  cents ;  then  we  divide  by 
6  to  find  the  interest  for  $8 ;  then  by  adding  both 
we  find  the  interest  for  $48,  as  required. 

EXAMPLE.  4 — Required  the  interest  of  $260  for 
104  days  at  7  per  cent. 


Aiis. 


104=the  days. 

04  cts=the  int.  of  $52  for  104  days. 
20  cts.     Multiply  by  5. 


Here  we  first  find  the  interest  of  $52,  because 
the  days  is  the  interest  in  cents ;  then  we  multiply 
by  5  to  get  it  for  $260.  We  could  have  worked 
this  note  by  the  Bankers'  Method,  just  as  well,  by 
cutting  off  two  figures  in  the  principal,  making 
$2.60  cts.  the  interest  for  52  days,  and  then  multi- 
ply by  2  to  get  it  for  104  days.  GShe  student  must 
remember  that  the  interest  of  $260  for  104  days  is 
just  the  same  as  the  interest  of  $104  for  260  days. 


BANKERS'  METHOD  OF  COMPUTING  INTEREST.  107 

Problems  Solved  by  Both  Methods. 
We  will  now  solve  some  examples  by  both  meth- 
ods, to  further  illustrate  this  system,  and  for  the 
purpose  of  teaching  the  pupil  how  to  use  his  judg- 
ment. He  will  then  have  learned  a  rule  more  val- 
uable than  all  others. 

EXAMPLE  5.— What  is  the  interest  $180  for  75 
days  at  6  per  cent.? 

Operation  by  taking  advantage  of  the  $. 
75=the  days.  $60X3— $130. 

75  cts.=the  int.  of  $60  for  75  days. 
3  Multiply  by  3. 


Ans.   $2  |  25  cts.=the  int.  of  $180  for  75  days. 
Operation  by  the  Bankers'  Method. 
$180— the  principal.  60da.-f  15da.=75da. 

4)$1     80  cts.^the  int.  for  60  days. 
45  cts.~the  int.  for  15  days. 

Ans.  $2  |  25  cts.— the  int.  for  75  days. 

By  the  first  method  we  multiplied  by  3,  because 
3X$60=$180;  by  the  second  method  we  added 
on  J,  because  60da.-|-640da.:=75da. 

N.  B. — When  advantage  can  be  taken  of  both 
time  and  principal,  if  the  student  wishes  to  prove 
his  work,  he  can  first  work  it  by  the  Bankers' 
Method,  and  then  by  taking  advantage  of  the  prin- 
cipal, or  vice  versa.  And  as  the  two  operations  are 
entirely  different,  if  the  same  result  is  obtained  by 
each,  he  may  fairly  conclude  that  the  work  is  correct 


108       ORTON'S  LIGHTNING  CALCULATOR. 

LIGHTNING  METHOD 
OP 

COMPUTING  INTEREST 

On  all  notes  that  bear  $12  per  annum,  or  any  all* 
quot  part  or  multiple  of  $12. 

IF  a  note  bears  $12  per  annum,  it  will  certainly 
bear  $1  per  month;  hence  the  time  in  montha 
would  be  the  interest  in  $ ;  and  the  decimal  parts 
of  a  month  would  be  the  interest  in  decimal 
parts  of  a  $;  therefore  when  the  note  bears  $12 
per  annum  we  have  the  following  rule : 

RULE. — Reduce  the  years  to  months,  add  in  the 
given  months,  and  place  one-third  of  the  days  to  the 
right  of  this  number,  and  you  have  the  interest  in 
dimes. 

EXAMPLE  1. — Required  the  interest  of  $200 
for  3  years,  7  months,  and  12  days,  at  6  per  cent. 

200  J  of  12  days=4. 

6 

-  Yh.  Mo.  Da. 

$12.00z=int.  for  I  yr.  37  12=:43.4mo. 

Hence  43.4  dimes,  or  $43.40cts.,  Ans. 

We  see  by  inspection  that  this  note  bears  $12 
in**rest  a  year ;  hence  the  time  reduced  to  months, 


LIGHTNING  METHOD  OF  COMPUTING  INT.     105 

with  one-third  of  the  days  to  the  right,  is  the  in- 
terest in  dimes.  If  this  note  bore  $6  a  year,  in- 
stead of  $12,  we  would  take  one-half  of  the  above 
interest ;  if  it  bore  $18,  instead  of  $12,  we  would 
add  one-half;  if  it  bore  $24,  instead  of  $12,  we 
would  multiply  by  2,  etc. 

EXAMPLE  2. — Required   the  interest  of  $150 
for  2  years,  5  months,  and  13  days,  at  8  per  cent. 
150  J  of  13  days=4J. 

8 

Tr.  Mo.  Da. 

$12.00=int.  for  1  yr.  25  13=r29.4Jmos» 

Hence  $29.4J  dimes,  or  $29.43J  cts.,  Ans. 
We  see  by  inspection  that  this  note  bears  $12 
interest  a  year ;  hence  the  time  reduced  to  months, 
with  one -third  of  the  days  placed  to  the  right,  gives 
the  interest  at  once. 

EXAMPLE  3. — Required  the  interest  of  $160  for 
11  years,  11  months,  and  11  days,  at  7J  per  cent. 
160  J  of  11  days=3f. 

1% 

$12.00=int.  for  1  yr.  11  11  1  i=143.3|mo8. 

Hence  $143.3f  dimes,  or  $143.36f  cts.,  Ans. 

When  the  Interest  is  more  or  less  than  $12  a  Tear. 
RULE. — First  find  the  interest  for  the  given  time 
on  the  base  of  $12  interest  a  year  ;  then,  if  the  in- 
terest on  the  note  is  only  $6  a  year}  divide  by  2  j  if 


110          ORTON^S   LIGHTNING   CALCULATOR 

$24  a  year,  multiply  1y  2 ;  if  $18  a  year,  ado*  on 
one-half,  etc. 

EXAMPLE  1. — What  is  the  interest  of  $300  for 
4  years,  7  months,  and  18  days,  at  6  per  cent. 

J  of  18  days=6. 

300  4yr.  7mo.  18da.=55.6mo. 

6 

$i8.00:=int.  for  1  year.     2)55.6,  int.  at  $12  a  year. 
»18=li  times  $12.  278 

$83.4  An*. 

If  tne  interest  was  $12  a  year,  $55.60  would  be 
the  answer;  because  55.6  is  the  time  reduced  to 
months ;  but  it  bears  $18  a  year,  or  1  \  times  12 ; 
hence  1J  times  55.6  gives  the  interest  at  once. 

EXAMPLE  2. — .Required  the  interest  of  $150 
for  3  years,  9  months,  and  27  days,  at  4  per  cent. 

£  of  27  days=9. 

150  Syr.  9mo.  27da.r=45.9mo. 

4  2)45.9,  int.  at  $12  a  year. 


$6.00z=int.  for  1  year.     $22.95  Ans. 
$6:=£  times  $12. 

If  the  interest  was  $12  a  year,  $45.90  would  be 
the  answer ;  because  245.9  is  the  time  reduced  to 
months;  out  it  bears  $6  a  year,  or  \  times  12 j 
hence  \  nines  45.9  g'ves  the  interest  at  once. 


MERCHANTS    METHOD  OF  COMPUTING  xNT.     Ill 

« 

MERCHANTS'  METHOD 

OF 
COMPUTING    INTEREST. 

FOR  YEARS,  MONTHS,  AND  DAYS. 

THE  computation  of  simple  interest,  where  the 
lime  consists  of  years,  months,  and  days,  is  quite 
difficult.     Taking  the  aliquot  parts  for  the  differ- 
ent portions  of  time  almost  invariably  involves  the 
calculator  in  fractions,  and,  unless  he  is  well  versed 
in  vulgar  fract'  )ns  he  will  not  be  able  to  arrive  at 
the  correct  result.     We  have  three  bases  by  which 
sve  compute  interest    at   different  rates  per  cent, 
and  by  which  we  are  enabled  to  entirely  avoid  the 
use  of  fractions.    These  three  bases  are  each  obtained 
different  from  the  other,  and  consequently  we  have 
three  rules  for  computing  interest :  one  at  a  base  of 
sne  per  cent.,  a  second  at  a  base  of  twelve  per 
\ent.,  and  a  third  at  a  base  of  thirty-six  per  cent 
RULE  for  computing  interest  at  1  per  cent. : 
Take  one-third  of  the  number  of  days  and  annex 
o  the  number  of  months ;  divide  the  number  thu$ 
formed  by  12  ;  annex  the  quotient  thus  obtained  to 
ihe  number  of  years,  and  multiply  the  principal  by 
(his  number ;  if  the  principal  contains  cents,  point 
iff  five  decimal  places  ;  if  not,  point  off  three  deci- 
K 


112  ORTONS'  LIGHTNING  CALCULATOR. 

• 

mal  places;  this  will  give  the  interest  at  one  per 
cent.  For  any  other  rate  per  cent.,  multiply  the  in- 
terest at  one  per  cent,  by  the  required  rate  per  cent. 

Remark. — This  rule  applies  to  all  problems  in 
interest  where  the  days  are  divisible  by  3,  and  this 
number,  annexed  to  the  number  of  months,  divisi- 
ble by  12. 

EXAMPLE. 

Required  the  interest  on  $112,  at  1  per  cent.,  for 
3  years,  3  months  and  18  days. 

SOLUTION. 

Take  one-third  of  the  number  of  days,  J  of  18 
:nr6,  annex  this  number  to  the  months  given,  36, 
divide  this  number  by  12,  36-^-12=3,  annex  this 
number  to  the  year  gives,  33,  multiply  the  princi- 
pal by  33,  $112X33=3.69  6,  point  off  three  deci- 
mal places,  and  we  have  the  required  interest, 
$3.696. 

EXAMPLE. 

Required  the  interest  on  $125  12,  at  7  per  cent., 
for  2  years,  8  months  and  12  days. 

SOLUTION. 

Take  one-third  of  the  number  of  days,  $ 
of  12rr:4,  annex  this  number  to  the  number  of 
months  we  have  84,  divide  this  number  by  12, 


MERCHANTS'  METHOD  OF  COMPUTING  INT.   113 

34-7-12=7,  annex  this  number  to  the 

number  of  years'  we  have  27,  multiply 

the    principal    by   this    number,  and 

point  off  five  decimal  places,  and  you 

have  the  interest  at  one  per  cent.;  mul- 

tiply this  interest  by  7,  and  you  have     -- 

the  interest  at  7  per  cent.,  the  required  $23  .64768 

rate. 

EXAMPLE. 

Kequired  the  interest  on  $1,023,  at  8  per  cent., 
for  1  year,  9  months  and  18  days. 

SOLUTION. 

Take  one-third  the  number  of  days  and  annex 
to  the  number  of  months,  \  of  18=6,  we  have 
96-r-12=8,  annex  this  number  to  the  years  $1023 
we  have  18,  multiply  the  principal  by  18 

this  number,  and  point  of  three  decimal  - 
places,  which  gives  the  interest  at  1  per  $18  .414 
cent.;  multiply  the  interest  at  one  per  8 

cent,  by  8,  and  you  have  the  required  in  -- 
terest.  $147  .312 


.  —  This  rule  will  apply  to  all  problems 
in  interest  if  one-third  of  the  number  of  the  days 
be  taken  decimally  and  annexed  to  the  number  of 
months,  and  this  number,  divided  by  12,  carried 
out  decimally.  But  this  makes  the  multiplier 
very  large  ;  hence,  to  avoid  this  large  number  in 


114         ORTON'S  LIGHTNING  CALCULATOR. 

the  multiplier,  where  the  days  are  divisible  by  $y 
and  this  number,  annexed  to  the  months,  is  uot 
divisible  by  12,  we  use  the  following  rule,  called 
our  base  at  12  per  cent. : 

RULE. — Reduce  the  years  to  months^  add  in  the 
months,  take  one-third  of  the  number  of  days  and 
annex  to  this  number,  multiply  the  principal  by  the 
number  thus  formed;  if  there  are  cents  in  the  prin- 
cipal, point  off  five  decimal  places  ;  if  there  are  no 
cents  in  the  principal,  point  off  three  decimal  places  ; 
this  gives  the  interest  at  12  per  cent.  For  any  other 
rate  per  cent.,  take  such  part  of  the  base  before  mul- 
tiplying as  the  required  rate  is  a  part  of  12. 

EXAMPLE. 

Required  the  interest  on  $123,  at  12  per  cent., 
for  2  years,  2  months  and  six  days. 

SOLUTION. 

Reduce   the    2   years   to   months   gives   us   24 
months,  add  on  the  2  months  gives  us  26 
months,  take  one-third  of  the  days,  J  of        $123 
6=2,  annexed   to   the  26  months    gives  262 

262,  which  constitutes  the  base ;  multiply    

the  principal  by  this  base,  and  you  have  $32  .226 
the  interest  at  12  per  cent. 

EXAMPLE. 

Required  the  interest  on  $144,  at  6  per  cent.,  for 
4  years,  5  months  and  12  days. 


MERCHANTS'  METHOD  OF  COMPUTING  INT.    115 

SOLUTION. 

Reduce  the  4  years  to  months  gives  48  mcntha, 
add  in  the  5  months  gives  53  months,  take  one- 
third  of  the  days  and  annex  to  the  number  of 
months,  J  of  12=4.  annex  to  the  53  months,  534 ; 
this  number  multiplied  into  the  principal  would 
give  the  interest  at  12  per  cent.  But  we  want  it 
at  6  per  cent.  We  will  now  take  such  part  of 
either  principal  or  base  as  6  is  a  part  of  12 ;  6  is 
|  of  12,  therefore  we  will  take  J  of  144—72 
one-half  of  the  principal,  and  mul-  534 

tiply  it   by  the  base,  which  will 

give  the  interest  at  6  per  cent.  $38.448 

EXAMPLE. 

Required  the  interest  on  $347  25,  at  8  per  cent., 
for  2  years,  3  months  and  9  days. 
SOLUTION. 

Reduce  the  2  years  to  months,  24  months,  add 
the  3  months,  27  months,  take  one-third  of  the 
days,  J  of  9=3,  annex  to  the  months,  273,  the 
base;  this,  multiplied  into  the  principal,  would 
give  the  interest  at  12  per  cent.  But  we  want  the 
interest  at  8  per  cent ;  we  will  take 
two-thirds  of  the  base  before  multiply-  $347  25 
ing:  f  of  273=182;  the  principal  182 

multiplied  by  this  number  gives  the 

interest  at  8  per  cent.  $63.19950 

Remark. — This  base  is  used  where  the  days  are 
divisible  by  3,  and  the  number  formed  by  annex- 
8 


1 1 0         ORTON'S  LIGHTNING  CALCULATOR. 

ing  one- third  of  the  days  to  the  months  not  divis)« 
ble  by  12.  We  now  come  to  time  in  which  neithei 
days  nor  months  are  divisible.  .Where  such  time 
as  this  occurs,  we  use  a  base  at  36  per  cent. 

RULE. — Reduce  the  time  to  days,  by  multiplying 
the  years  by  12,  adding  in  the  months,  if  any,  and 
multiplying  this  number  by  30,  adding  in  the  days, 
if  any;  multiply  the  principal  by  this  number , 
pointing  off  5  decimal  places,  where  cents  are  given 
in  the  principal,  and  3  places  where  no  cents  arc 
given.  This  will  give  the  interest  at  36  per  cent. 

EXAMPLE. 

Required  the  interest  on  $144,  at  36  per  cent., 
for  3  years,  2  months  and  2  days. 

SOLUTION. 

Reduce  the  time  to  days  gives  1142  $144 
days;  multiply  the  principal  by  this  base,  1142 

and   you  have   the   interest    at   36    per  : — - 

cent  $164.448 

EXAMPLE. 

Required  the  interest  on  $144,  at  9  per  cent.,  ror 
5  years,  7  months  and  5  days. 
SOLUTION. 

Reduce  the  time  to  days  gives  2,015  days ;  if 
we  multiply  the  principal  by  this  base,  we  would 
get  the  interest  at  36  per  cent.;  but  we  want  it  at 
9  per  cent.  We  can  take  such  part  of  either 


MERCHANTS'  METHOD  OF  COMPUTING  INT.  117 

principal  or  base  as  9  is  a  part  of  36  before  multi- 
plying ;  9  is  J  of  36 ;  we  will  take  J-  of  the  prin- 
cipal, it  being  divisible  by  4 ;  J  of  144=36,  2915 
which,  multiplied  into  the  base,  will  give  35 
the  interest  at  9  per  cent.,  by  pointing  off 

3  decimal  places.  $72.540 

EXAMPLE. 

Required  the  interest  on  $875  15,  at  6  per  cent., 
for  5  years,  7  months  and  12  days. 
SOLUTION. 

Reduce  the  time  to  days  gives  2022  days ;  6  is 
J  of  36 ;  take  one  sixth  of  the  base, 
£  of  2022=337;  multiply  the  prin-         $875  15 
cipal  by  this  number,  point  off  5  dec-  337 

imal  places,  and  you  have  the  interest 

at  6  per  cent.,  the  required  rate.  $294.92555 

Remark. — We  have  now  fully  explained  our 
method  of  computing  interest  at  the  three  different 
bases.  Any  and  every  problem  in  interest  can  be 
solved  by  one  of  these  three  bases.  Some  prob- 
lems can  be  solved  easier  by  one  base  than  another. 
Where  the  days  are  divisible  by  3,  and  their  num- 
ber, annexed  to  the  months,  divisible  by  12,  it  is 
the  shortest  and  best  method  to  use  the  base  at  1 
per  cent.  By  using  one  or  the  other  of  these  three 
bases,  the  student  can  avoid  the  use  of  vulgar 
fractions.  The  student  must  study  these  thi»ee 
principles  carefully,  and  learn  to  adopt  readily  the 
base  best  suited  to  the  problem  to  be  solved. 


118         ORTON'S  LIGHTING  CALCULATOR. 
PARTIAL  PAYMENTS 

ON   NOTES,    BONDS,   AND   MORTGAGES. 

Ta  compute  interest  on  notes,  bonds,  and  mort- 
gages, on  which  partial  payments  have  been  made, 
two  or  three  rules  are  given.  The  following  is 
called  the  common  rule,  and  applies  to  cases  where 
the  time  is  short,  and  payments  made  within  a  year 
of  each  other.  This  rule  is  sanctioned  by  custom 
and  common  law ;  it  is  true  to  the  principles  of 
simple  interest,  and  requires  no  special  enactment. 
The  other  rules  are  rules  of  law,  made  to  suit  such 
cases  as  require  (either  expressed  or  implied)  an- 
nual interest  to  be  paid,  and  of  course  apply  to  no 
business  transactions  closed  within  a  year. 

RULE. —  Compute  the  interest  of  the  principal  sum 
for  the  whole  time  to  the  day  of  settlement,  and  find 
the  amount.  Compute  the  interest  on  the  several  pay- 
ments, from  the  time  each  was  paid  to  the  day  of 
settlement;  add  the  several  payments  and  the  inter- 
est on  each  together,  and  call  the  sum  the  amount  of 
the  payments.  Subtract  the  amount  of  the  payment! 
from  the  amount  of  the  principal,  will  leave  the  sum 
due. 


PARTIAL   PAYMENTS.  119 

EXAMPLES. 

1.  A  gave  his  note  to  B  for  $10,000 ;  at  the  end 
of  4  months,  A  paid  $6,000 ;  and  at  the  expiration 
of  another  4  months,  he  paid  an  additional  sum  of 
$3,000 ;  how  much  did  he  owe  B  at  the  close  of  the 
year? 

By  the  Common  Rule. 

Principal $10,000 

Interest  for  the  whole  time 600 


Amount $10,600 

1st  payment $6,000 

Interest,  8  months      240 

2d  payment 3,000 

Interest,  4  months        60 


Amount $9,300  9,300 

Due $1300 

PROBLEMS   IN   INTEREST. 

There  are  four   parts  or  quantities  connected 
with  each  operation  in  interest:  these  are,  the 
Principal,  Hate  per  cent.,  Time,  Interest  or  Amount. 

If  any  three  of  them  ar3  given  the  other  may  be 
found. 

Principal,  interest,  and  time  given,  to  find  the 
rate  per  cent. 

1.  At  what  rate  per  cent,  must  $500  be  Dut  0* 
interest  to  gain  $120  in  4  years? 


[20       ORTON'S  LIGHTNING  CALCULATOR. 

Operation.  By  analysis. 

$500  The   interest  of 

.01  $1    for    the    given 

time  at  1  per  cent. 

5.00  is  4   cents.      $500 

4  will    be  500  times 

asmuch— 500X-04 

2ft. 00)120.00(6  per  cent.,  Ans.  =$20.00.    Then  if 
120.00  $20  give  1  per  cent., 

$120  will  give  \$> 
=6  per  cent. 

RFLE. — Divide  the  given  interest  "by  the  interest 

of  the  given  sum  at   1  per  cent,  for  the  given  timey 

and  the  quotient  'will  be  the  rate  per  cent,  required 

Principal,  interest,  and  rate  per  cent,  given,  to 

find  the  time. 

2.  How  long  must  $500  be  on  interest  at  6  per 
cent,  to  gain  $120  ? 

Operation  By  analysis. 

$500  We  find  the  in- 

.06  terest  of  $1.00    at 

the  given  rate   for 

30.00)120.00(4  years,  Ans.        1   year  is  6  cents 
120.00  $500,  will  therefore 

be    500    times    as 

much=500X  .06=$30.00.  Now,  if  it  take  1  year 
to  gain  $30,  it  will  require  yy>  to  gain  f  120=4 
years,  Ans. 


PARTIAL   PAYMENTS.  121 

RULE. — Divide  the  given  interest  by  the  interest 
of  the  principal  for  1  year,  and  the  quotient  is  the 
time. 

Given  the  amount,  time,  and  rate  per  cent.,  to 
find  the  principal. 

RULE. — Divide  the  given  amount  by  the  amount 
0/*$l,  at  the  given  rate  per  cent.,  for  the  given  time. 

REMARK. — This  rule  is  deduced  from  the  fact 
that  the  amount  of  different  principals  for  the  same 
time  and  at  the  same  rate  per  cent.,  are  to  each 
other  as  those  principals. 

BANK   DISCOUNT. 

Bank  Discount  is  the  sum  paid  to  a  hank  for  the 
payment  of  a  note  before  it  becomes  due. 

The  amount  named  in  a  note  is  called  the  face 
of  the  note.  The  discount  is  the  interest  on  the 
face  of  the  note  for  3  days  more  than  the  time 
specified,  and  is  paid  in  advance.  These  3  days 
are  called  days  of  grace,  as  the  borrower  is  not 
obliged  to  make  payment  until  their  expiration. 
Hence,  to  compute  bank  discount,  we  have  the  fol- 
lowing 

RULE. — Find  the  interest  on  the  face  of  the  note 
for  3  days  more  than  the  TIME  specified ;  this  will 
be  the  discount.  From  the  face  of  the  note  deduct 
the  discount,  and  the  remainder  will  be  the  PRESENT 
VALUE  of  the  note. 


122        ORTON'S  LIGHTNING  CALCULATOR. 

DISCOUNT,  OR  COUNTING  BACK. 
The  object  of  discount  is  to  show  us  what  al- 
lowance should  be  made  when  any  sum  of  money 
is  paid  before  it  becomes  due. 

The  present  worth  of  any  sum  is  the  principal 
that  must  be  put  at  interest  to  amount  to  that  sum  ^ 
in  the  given  time.  That  is,  $100  is  the  present  worth 
of  $106  due  one  year  hence;  because  $100  at  6  per 
cent,  will  amount  to  $106 ;  and  $6  is  the  discount. 
1.  What  is  the  present  worth  of  $12.72  due  one 
year  hence  ? 

First  method.  Second  method. 

$12.72  $ 

100  1.06)12.72($12  Ans. 

10.6 

106)1272.00($12  Am. 

106  2.12 

2.12 

212  

212 


As  $100  will  amount  to  $106  in  one  year  at  6 
per  cent.,  it  is  evident  that  if  {{Jg  of  any  sum  be 
taken,  it  will  be  its  present  worth  for  one  year,  and 
that  T$g  will  be  the  discount.  And  as  $1  is  the 
present  worth  of  $1.06  due  one  year  hence,  it  is 
evident  that  the  present  worth  of  $12.72  must  be 
equal  to  the  number  of  times  $12.72  will  contain 
81.06. 


EQUATION  OF  PAYMENTS.  123 

RULE. — Divide  the  given  sum  by  the  amount  of 
81  for  the  given  rate  and  time,  and  the  quotient  will 
be  the  present  worth.  If  the  present  worth  be  sub- 
tracted from  the  given  sum,  the  remainder  will  be  the 
discount. 


EQUATION  OF  PAYMENTS. 

EQUATION  OF  PAYMENTS  is  the  process  of  find- 
ing the  equalized  or  average  time  for  the  payment 
of  several  sums  due  at  different  times,  without  loss 
to  either  party. 

To  find  the  average  or  mean  time  of  payment, 
when  the  several  sums  have  the  same  date. 

RULE. — Multiply  each  payment  by  the  time  that 
must  elapse  before  it  becomes  due;  then  divide  the 
sum  of  these  products  by  the  sum  of  the  payments, 
and  the  quotient  will  be  the  averaged  time  required. 

NOTE. — When  a  payment  is  to  be  made  down,  it 
has  no  product,  but  it  must  be  added  with  the 
other  payments  in  finding  the  average  time. 

EXAMPLE  1. — I  purchased  goods  to  the  amount 
of  $1200;  $300  of  which  I  am  to  pay  in  4  months 
$100  in  5  months,  and  $500  in  8  months.  How 
long  a  credit  ought  I  to  receive,  if  I  pay  tho 
wholo  sum  at  once?  Ans.  6  months. 

L 


124          ORION'S   LIGHTNING  *  CALCULATOR. 

Mo.  Mo.  f    A  credit  on  $300  for  4  months  if 

><<v/OAA       1  OAA  -<  the  same  as  the  credit  on  $1  for 

4X  O  00=1^00  1  !200  months. 

f    A  credit  on  $400  for  5  months  ii 

5X400—2000  -{the  same  as  the  credit  on  $1  for 

(2000  months. 

Qs/  P\fifi AH fifi  I      A  credit  on  $500  for  8  months  is 

OX^UV ^UUU  J  the  same  as  the  credit  on  $1  for 

.      (^000  months. 

1  OAA\  rroAA  sc*  Therefore,  I    should   have    the 

1200)  7ZOO  (b  mO.       same  credit  as  a  credit  on  $1  for 

-9n  A  7200    months,   and  on  $1200,  the 

'  ^vJU  whole  sum,  one-twelfth  hundredth 

_____  part  of  7200  monthi,  which  is  6 

months. 

This  rule  is  the  one  usually  adopted  by  mer- 
chants, although  not  strictly  correct,  still,  it  is  suf- 
ficiently accurate  for  all  practical  purposes* 

To  find  the  average  or  mean  time  of  payment, 
when  the  several  sums  have  different  dates. 

EXAMPLE  1. — Purchased  of  James  Brown,  at 
pundry  times,  and  on  various  terms  of  credit,  as  by 
the  statement  annexed.  When  is  the  medium  time 
of  payment? 

Jan.      1,  a  bill  am'ting  to  $360,  on  3  months'  credit. 
Jan.    15,  do.       do.  186,  on  4  months' credit. 

March  1,  do.       do.  450,  on  4  months'  credit. 

May    15,  do.       do.  300,  on  3  months' credit 

June  20,  do.       do.  500,  on  5  months' credit, 

Ans.  July  24th,  or  in  115  da. 
Due  April    1,  $360 

May   15,    186X  44=    8184 

July     1,    450X   91=  40950 

Aug.  15,    300Xl36=r  40800 

Nov.  20,    500X233=116500 

1796V  into  )206434(114|||  days. 


EQUATION    OF   PAYMENTS.  125 

We  first  find  the  time  when  each  of  the  billa 
will  become  due.  Then,  since  it  will  shorten  the 
operation  and  bring  the  same  result,  we  take  the 
time  when  the  first  bill  becomes  due,  instead  of  its 
date,  for  the  period  from  which  to  compute  the 
average  time.  Now,  since  April  1  is  the  period 
from  which  the  average  time  is  computed,  no  time 
will  be  reckoned  on  the  first  bill,  but  the  time  for 
the  payment  of  the  second  bill  extends  44  days  be- 
yond April  1,  and  we  multiply  it  by  44. 

Proceeding  in  the  same  manner  with  the  remain- 
ing bills,  we  find  the  average  time  of  payment  to 
be  114  days  and  a  fraction,  from  April  1,  or  on  the 
24th  of  July. 

RULE. — Find  the  time  when  each  of  the  sums  be- 
comes due,  and  multiply  each  sum  by  the  number  of 
days  from  the  time  of  the  earliest  payment  to  the 
payment  of  each  sum  respectively.  Then  proceed  as 
,in  the  last  rule,  and  the  quotient 'will  be  the  aver- 
age time  required,  in  days,  from  .  the  earliest  pay- 
ment. 

NOTE. — Nearly  the  same  result  may  be  obtained 
by  reckoning  the  time  in  months. 

In  mercantile  transactions  it  is  customary  to  give 
a  credit  of  from  3  to  9  months,  on  bills  of  sale. 
Merchants  in  settling  such  accounts,  as  consist  of 
various  items  <*f  debit  and  credit  for  different  times, 
generally  employ  the  following : 


1  26  ORTON  S  LIGHTNING  CALCULATOR. 

RULE. — Place  on  the  debtor  or  credit  side,  such  a 
mm,  (which  may  be  called  MERCHANDISE  BALANCE,) 
as  will  balance  the  account. 

Multiply  the  number  of  dollars  in  each  entry  by 
the  number  of  days  from  the  time  the  entry  was  made 
to  the  time  of  settlement;  and  the  Merchandise  bal- 
ance by  the  number  of  days  for  which  credit  was 
given.  Then  multiply  the  difference  between  the  sum 
of  the  debit,  and  the  sum  of  the  credit  products,  by 
the  interest  of  $1  for  1  day ;  this  product  will  be 
the  INTEREST  BALANCE. 

When  the  sum  of  the  debit  products  exceed  the  sum 
of  the  credit  products,  the  interest  balance  is  in  favor 
of  the  debit  side ;  but  when  the  sum  of  the  credit 
products  exceed  the  sum  of  the  debit  products,  it  is  in 
favor  of  the  credit  side.  Now  to  the  merchandise 
balance  add  the  interest  balance,  or  subtract  it,  as  the 
case  may  require,  and  you  obtain  tlie  CASH  BAL- 
ANCE. 

A  has  with  B  the  following  account : 

1849.  Dr.    I      1849.  dr. 

Jan.  2.     To  merchandise,    $200      Feb.   20.  By  merchandise,   $10( 
April  20.  "  "  400 1    May.  10.  "  30C 

If  interest  is  estimated  at  7  per  cent.,  and  a 
credit  of  GO  days  is  allowed  on  the  different  sums, 
what  is  the  cash  balance  August  20,  1849? 

Ans.  206.54. 

EXPLANATION. — Without  interest  the  cash  bal 
ance  would  be  $200. 


EQUATION   OF   PAYMENTS  127 

If  «io  Credit  had  been  given,  the  debits  should 
oo  increased  by  the  interest  of  $200  for  230  days, 
at  7  percent.;  and  the  interest  of  $400  for  122 
days,  at  7  per  cent.  The  credits  should  be  increas- 
ed by  the  interest  of  $100  for  181  days,  at  7  per 
cent.,  and  the  interest  of  $300  for  102  days,  at  7 
per  cent. 

Since  a  credit  of  60  days  is  given  on  all  sums,  it 
is  evident  by  the  above  calculation,  that  we  should 
increase  the  debits  by  the  interest  of  the  sum  of 
the  debits,  $600,  for  60  days  more  than  justice  re  • 
quires.  Also,  that  we  should  increase  the  credits 
by  the  interest  of  the  sum  of  the  credits,  $400,  for 
60  days  more  than  we  should  do. 

Now,  instead  of  deducting  these  items  of  inter- 
est from  the  amount  of  debit  and  credit  interests, 
it  is  plain  that  it  will  be  more  convenient  and 
equally  just,  to  diminish  the  debit  interest  of  the 
merchandise  balance  for  60  days,  which  can  be 
most  readily  accomplished  by  adding  the  interest 
on  the  merchandise  balance  for  60  days,  to  the 
credit  items  of  interest. 

|  From  which  we  discover  that  the  interest  balance 
is  equal  to  the  difference  between  the  sum  of  the 
debit  interests,  and  the  sum  of  the  credit  interests 
increased  by  the  interest  of  the  merchandise  bal- 
ance for  the  time  for  which  Credit  was  given. 


128        OETON'S  LIGHTNING  CALCULATOB. 
Operation. 

DEBITS.  CREDITS. 

200X230—46000  100x181=18100 

400X122=48800  300x102=30600 

•  Balance,  200X   60=12000 


94800 
60700 


60700 


0.07 


X34100=$6.54  Interest  balance, yearly 

365 

Therefore,  the  foregoing  account  becomes  bal- 
anced as  follows : 


1849.  Dr. 

Jan.     2.  To  Merchandise,  $200.00 
April 20.  "  "  400.00 

Aug.  20.  "balance  of  int.       6.54 


Ang.  20. 


$606.54 
1  Cash  balance,  $206.54 


1849  Or. 

Feb.  20.  By  Merchandise,  $100.00 
May.  10.  "  "  300.00 

Aug.  20.  "  balance,  206.54 


$606.54 


NOTE. — It  is  customary  in  practice,  when  the 
number  of  cents  in  any  of  the  entries,  are  less  than 
DO,  to  omit  them,  and  to  add  $1  when  they  are  50, 
or  more. 


SQUARE  AND  CUBE  BOOTS. 

SQUARE  AND  CUBE  ROOTS, 

To  work  the  square  and  cube  roots  with  ease  and 
facility,  the  pupil  must  be  familiar  with  the  follow 
ing  properties  of  numbers : 

Their  importance  can  not  be  exaggerated  if  we 
wish  to  insure  skill  or  even  sound  information  on 
this  subject. 

I.  A  square  number,  multiplied   by   a  square 
number,  the  product  will  be  a  square  number. 

II.  A  square  number,  divided  by  a  square  num- 
ber, the  quotient  is  a  square. 

III.  A  cube  number,  multiplied  by  a  cube,  the 
product  is  a  cube. 

IY.  A  cube  number,  divided  by  a  cube,  the  quo- 
tient will  be  a  cube. 

V.  If  the  square  root  of  a  number  is  a  compos- 
ite number,  the   square  itself  may  be  divided  intv 
integer  square  factors ;  but  if  the  root  is  a  prime 
number,  the  square  can  not  be  separated  into  square 
factors  without  fractions. 

VI.  If  the  unit  figure  of  a  square  number  is  5,  we 
may  multiply  by  the  square  number  4,  and  we  shall 
have  another  square,  whose  unit  period  will  be 
ciphers. 

VII.  If  the  unit  figure  of  a  cube  is  5,  we  may 
multiply  by  the  cube  number  8,  and  produce  an- 
other  cube,  whose  unit  period  will  be  ciphers. 


130 


ORTON'S  LIGHTNING  CALCULATOR 


K.  B.  If  a  supposed  cube,  whose  unit  figure  is 
5,  be  multiplied  by  8,  and  the  product  does  not  give 
three  ciphers  on  the  right,  the  number  is  not  a  cube. 

We  present  the  following  table,  for  the  pupil  to 
compare  the  natural  numbers  with  the  unit  figure 
of  their  squares  and  cubes,  that  he  may  be  able  to 
extract  roots  by  inspection. 


Numbers  

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

Squares  

1 

4 

9 

16 

25 

36 

49 

C4 

81 

100 

Cubes  

1 

8 

?7 

64 

125 

fllfi 

343 

51ft 

7ft9 

1000 

EXERCISES  FOR  PRACTICE. 

1.  What  is  the  square  root  of  625?     Ans.  25 
If  the  root  is  an  integer  number,  we  may  know, 

by  the  inspection  of  the  table,  that  it  must  be  25 , 
as  the  greatest  square  in  6  is  2,  and  5  is  the  only 
figure  whose  square  is  5  in  its  unit  place. 

Again,  take  625 

Multiply  by  44  being  a  square. 

2500 

The  square  root  of  this  product  is  obviously  50; 
but  this  must  be  divided  by  2,  the  square  root  of 
4,  which  gives  25,  the  root. 

2.  What  is  the  square  root  of  6561  ?     Ans.  81. 
As  the  unit  figure,  in  this  example,  is  1,  and  in 


SQUARE  AND  CUBE  ROOTS.  131 

the  line  of  squares  in  the  table,  we  find  1  only  at  1 
and  81,  we  will,  therefore,  divide  6561  by  81,  and  we 
find  the  quotient  81 ;  81  is,  therefore,  the  square  root. 

3.  What  is  the  square  root  of  106729?  Ans.  327. 
As  the  unit  figure,  in  this  example,  'is  9,  if  the 

number  is  a  square,  it  must  divide  by  either  9,  or 
49.  After  dividing  by  9  we  have  11881  for  the 
other  factor,  a  prime  number,  therefore  its  root  is  a 
prime  number=109.  109,  multiplied  by  3,  the 
root  of  9,  gives  327  for  the  answer. 

4.  What  is  the  root  of  451584?         Ans.  672. 
As  the  unit  figure  is  4,  and  in  the  line  of  squares 

we  find  4  only  at  4  and  64,  the  above  number,  if  a 
square,  must  divide  by  4,  or  64,  or  by  both. 

We  will  divide  it  by  4,  and  we  have  the  factors 
4  and  112896.  This  last  factor  closes  in  6  ;  there- 
fore, by  looking  at  the  table,  we  see  it  must  divide 
by  16,  or  36,  etc. 

We  divide  by  36,  and  we  have  the  factors  36  and 
3136;  divide  this  last  by  16,  and  we  have  16  and 
196 ;  divide  this  last  fraction  by  4,  and  we  have  4 
and  49. 

Take  now  our  divisors,  and  last  factor,  49,  and 
we  have  for  the  original  number  the  product  of 
4X36X16X^X49;  the  roots  of  which  are  2x6 
X4X2X?,  the  products  of  which  are  672,  the 
answer 

5.  Extract  the  square  root  of  2025.     Ans.  45. 

9 


132       ORTON'S  LIGHTNING*  CALCULATOR. 

1st.  Divide  by  the  square  number  25,  and  we 
find  the  two  factors,  25x81?  as  equivalent  to  the 
given  number.  Roots  of  these  factors,  5x9—45, 
the,  answer. 

Again,  multiply  by  the  square  number  4,  when 
a  number  ends  in  25,  and  we  have  8100,  root  90, 
half  of  which,  because  we  multiplied  by  4,  the 
square  of  2,  is  45,  the  answer. 

Problems  on  the  Right-angled  Triangle. 

1.  The  top  of  a  castle  is  45  yards  high,  and  is 
surrounded  with  a  ditch,  60  yards  wide ;  required 
the  length  of  a  ladder  that  will  reach  from  the 
outside  of  the  ditch  to  the  top  of  the  castle. 

Ans.  75  yards. 

This  is  almost  invariably  done  by  squaring  45 
and  60,  adding  them  together,  and  extracting  the 
square  root ;  but  so  much  labor  is  never  necessary 
when  the  numbers  have  a  common  divisor,  or  when 
the  side  sought  is  expressed  by  a  composite  number. 

Take  45  and  60;  both  may  be  divided  15,  and 
they  will  be  reduced  to  3  and  4.     Square  these, 
9+16=25.     The  square  root  of  25  is  5,  which, 
multiplied  by  15,  gives  75,  the  answer. 
Abbreviations  in  Cube  Root 

1.  WhaA,  is  the  cube  root  of  91125  ?     Ans.  45. 
Multiply  by  8 

729000 


SQUARE  AND  CUBE  ROOTS.        1?  j 

Now,  729  being  the  cube  of  9,  the  root  of 
T29000  is  90 ;  divide  this  by  2,  the  cube  root  of  8, 
and  we  have  45,  the  answer. 

When  it  is  requisite  to  multiply  several  numbers 
together  and  extract  the  cube  root  of  their  pro- 
duct, try  to  change  them  into  cube  factors  and  ex- 
tract the  root  before  multiplication. 

EXAMPLES. 

1.  What  is  the  side  of  a  cubical  mound  equal 
to  one  288  feet  long,  216  feet  broad,  and  48  feet 
high? 

The  common  way  of  doing  this,  is  to  multiply 
these  numbers  together  and  extract  the  root — a 
lengthy  operation.  But  observe  that  216  is  a  cube 
number,  and  288=2x12X12,  and  48=4x12; 
therefore  the  whole  product  is  216x8Xl2Xl2X 
12.  Now,  the  cube  root  of  216  is  6,  of  8  is  2,  and 
of  12s  is  12,  and  the  product  of  6X2X12—144, 
the  answer. 

2.  Required  the  cube  root  of  the  product  of 
448X392  the  short  way.  Ans.  56. 

We  can  extract  the  root  of  cube  numbers  by  in 
spection  when  they  do  not  contain  mwe  than  tw* 
periods. 


13£       ORTON'S  LIGHTNING  CALCULATOR. 

KULE. — As  there  will  be  two  figures  in  the  root,  the  firs* 
may  easily  be  found  mentally,  or  by  the  Table  of  Powers]  and 
if  the  unit  figure  of  the  power  be  1,  the  unit  figure  in  the  root 
will  be  1 ;  and  if  it  be  8,  the  root  will  be  two]  and  if  7,  it 
will  be  3 ;  and  if  the  unit  of  the  power  be  6,  the  unit  of  the 
root  will  be  6;  and  if  5,  it  will  be  5]  \f3,  it  will  be  7  ;  t/2, 
it  will  be  8;  and  if  the  unit  of  the  power  be  9,  the  unit  of  the 
root  will  be  9.  This  will  appear  evident  by  inspecting  the 
Table  of  Powers. 

EXAMPLES. 

Find  the  cube  root  of  195112.  This  number 
consists  of  two  periods.  Compare  the  superior 
period  with  the  cubes  in  the  table,  and  we  find  that 
195  lies  between  125  and  216.  The  cube  root  of 
the  tens,  then,  must  be  5.  The  unit  figure  of  the 
given  cube  is  2 ;  and  no  cube  in  the  table  has  2 
for  its  unit  figure,  except  512,  whose  root  is  8; 
therefore  58  is  the  root  required. 

What  is  the  cube  root  of  97336  ?        Ans  46. 

EXPLANATION. — By  examining  the  left  hand 
period,  we  find  the  root  of  97  is  4,  and  the  cube  of 
4  is  64.  The  root  can  not  be  5,  because  the  cube 
of  5  is  125.  The  unit  figure  of  the  given  cube  is 
6 ;  and  no  cube  in  the  table  has  6  for  its  unit  fig- 
ure, except  216,  whose  root  is  6  ;  the  answer,  there- 
tore,  is  46. 

The  number  912673  is  a  cube  ;  what  is  its  root? 

Ans.  97. 

Observe,  the  root  of  the  superior  period  must 


SQUARE  AND  CUBE  ROOTS        135 

be  9,  and  the  root  of  the  unit  period  must  be  some 
number  which  will  give  3  for  its  unit  figure  when 
cubed ;  and  7  is  the  only  figure  that  will  answer. 

The  following  numbers  are  cubes  j  required  their 
roots. 

1.  What  is  the  cube  root  of  59319?      Ans.  39 

2.  What  is  the  cube  root  of  79507?      Ans.  43. 

3.  What  is  the  cube  root  of  11^649  ?    Ans.  49. 

4.  What  is  the  cube  root  of  110592?    Ans.  48. 

5.  What  is  the  cube  root  of  357911  ?    Ans.  71. 

6.  What  is  the  cube  root  of  389017  ?    Ans.  73. 

7.  What  is  the  cube  root  of  571787  ?    Ans.  83. 
When  a  cube  has  more  than  two  periods,  it  can 

generally  be  reduced  to  two  by  dividing  by  some 
one  or  more  of  the  cube  numbers,  unless  the  root 
is  a  prime  number. 

The  number  4741632  is  a  cube;  required  its 
root.  Here  we  observe  that  the  unit  figure  is  2  ; 
the  unit  figure  of  the  root  must  therefore  be  the 
root  of  512,  as  that  is  the  only  cube  of  the  9  dig- 
its whose  unit  figure  is  2.  The  cube  root  of  512 
is  8 ;  therefore  8  is  the  unit  figure  in  the  root,  and 
the  root  is  an  even  number,  and  can  be  divided  by 
2 ;  and  of  course  the  cube  itself  can  be  divided  by 
8,  the  cube  of  2.  8)4741632 

592704 

Now,  as  the  first  number  was  a  cube,  and  being 
M 


136       ORTON'S  LIGHTNING  CALCULATOR. 

divided  by  a  cube,  the  number  592704  must  be  a 
cube,  and  by  inspection,  as  previously  explained, 
its  root  must  be  84,  which,  multiplied  by  2,  gives 
168,  the  root  required. 

The  number  13312053  is  a  cube;  what  is  ita 
root?  Ans.  237. 

As  there  are  three  periods,  there  must  be  three 
figures,  units,  tens,  and  hundreds,  in  the  root;  the 
hundreds  must  be  2,  the  units  must  be  7.  Let  us 
then  divide  the  2d  figure,  or  the  tens,  in  the  usual 
way,  and  we  have  237  for  the  root. 

Again,  divide  13312053  by  27,  and  we  have 
493039  for  another  factor.  The  root  of  this  last 
number  must  be  79,  which,  multiplied  by  3,  the 
cube  root  of  27,  gives  237,  as  before. 

The  number  18609625  is  a  cube;  what  is  its 
root? 

As  this  cube  ends  with  5,  we  will  multiply  ic 
by  8: 

18609625- 
8 


148877000 

As  the  first  is  a  cube,  this  product  must  be  a  cube; 
and  as  far  as  labor  is  concerned,  it  is  the  same  as 
reduced  to  two  periods,  and  the  root,  we  perceive 
at  once,  must  be  530,  which,  divided  by  2,  gives 
265  for  the  root  required. 


SQUARE   AND   CUBE   ROOTS.  137 

N.  B. — If  a  number,  whose  unit  figure  is  5,  be 
fflttltiplied  by  8,  and  does  not  result  in  three  ciphers 
on  the  right,  the  number  is  not  a  cube. 

To  find  the  Approximate  Cube  Root  of  Surds. 

RULE. —  Take  the  nearest  rational  cube  to  the  given 
number,  and  call  it  the  assumed  cube  ;  or  assume  a 
root  to  the  given  number  and  cube  it.  Double  the 
assumed  cube  and  add  the  number  to  it;  also  double 
the  number  and  add  the  assumed  cube  to  it.  Take 
the  difference  of  these  sums,  then  say,  As  double  of 
the  assumed  cube,  added  to  the  number,  is  to  this  dif- 
ference, so  is  the  assumed  root  to  a  correction. 

This  correction,  added  to  or  subtracted  from  the 
assumed  root,  as  the  case  may  require,  will  give  the 
cube  root  very  nearly. 

By  repeating  the   operation  with   the  root  last 
found  as  an  assumed  root,  we  may  obtain  results 
to  any  degree  of  exactness;  one  operation,  how 
ever,  is  generally  sufficient. 

EXAMPLES. 

1.  Required  the  cube  root  of  66. 

The  cube  root  of  64  is  4.  Now  it  is  manifest 
that  the  cube  root  of  66  is  a  little  more  than  4, 
and  by  taking  a  similar  proportion  to  the  preced- 
ing, we  have 

64x2  =  128     2*66=132 
66  64 

194  196:  :4  :  to  root  of  66. 


138       ORTON'S  LIGHTNING  CALCULATOR. 

Or,  194  :  2  :  :  4  :  to  a  correction 

194)8.0000(0.04124 
776 

240 
194 

460 

388 

720 

Therefore  the  cube  root  of  66  is  4.04124. 
2.  Required  the  cube  root  of  123. 
Suppose  it  5  ;  cube  it,  and  we  have  125. 
Now  we  perceive  that  the  cube  of  5  being 
than  123,  the  correction  for  5  must  be  subtracted. 

2X125=250     246 
Add  123     125 

As  373  :  371  :  :  5  :  root  of  123. 

Or,    373    :    2     :  :     5     :    correction  for  5 

373)10.0000(0.02681 
746 

2  540  From  5.00000 

2  238  Take  0.02681 


3020         Ans.  4.97319 

2984 

360. 


SQUARE   AND   CUBE   ROOTS.  139 

3.  What  is  the  cube  root  of  28  ?    Ans.  3,03658+ 

4.  What  is  the  cube  root  of  26  ?     Ans.  2,96249+ 

5.  What  is  the  cube  root  of  214  ?  Ans.  5,98142+ 
j     6.  What  is  the  cube  root  of  346  ?  Ans.  9,02034+ 

The  above  being  very  near  integral  cubes — that 
is.  28  and  26  are  both  near  the  cube  number  27, 
214  is  near  216,  etc.  All  numbers  very  near  cube 
numbers  are  easy  of  solution. 

We  now  give  other  examples,  more  distant  from 
integral  cubes,  to  show  that  the  labor  must  be  more 
lengthy  and  tedious,  though  the  operation  is  the 
same. 

1.  What  is  the  cube  root  of  3214?  Ans.  14,75758. 

Suppose  the  root  is  15 — its  cube  is  3375,  which, 
being  greater  than  3214,  shows  that  15  is  too  great ; 
the  correction  will  therefore  be  sub  tractive. 

By  the  rule,  9964  :  161  :  :  15.  0,243,  the 
correction. 

Assumed  root 15,0000 

Less...  2423 


Root  nearly....  14,7577 

Now  assume  14,7  for  the  root,  and  go  over  the 
operation  again,  and  you  will  have  the  true  root  to 
8  or  10  places  of  decimals. 

N.  B. — Roots  of  component  powers  may  be  ob- 
tained more  readily  thus : 

For  the  4th  root,  take  the  square  root  of  the 
square  root. 


140       ORTON'S  LIGHTNING  CALCULATOR. 
APPLICATION  OF  THE  CUBE  EOOT. 

PRINCIPLES   ASSUMED. 

Spheres  are  to  each  other  as  the,  cubes  of  their 
diameter. 

Cubes,  and  all  solids  whose  corresponding  parts 
are  similar  and  proportional  to  each  other,  are  to 
each  other  as  the  cubes  of  their  diameters,  or  of 
their  homologous  sides. 

1.  If  a  ball,  3  inches  in  diameter,  weigh  4  pounds, 
what  will  be  the  weight  of  a  ball  that  is  6  inches 
in  diameter?  Ans.  321bs. 

2.  If  a  globe   of  gold,  1   inch  in  diameter,  be 
worth  $120,  what  is  the  value  of  a  globe,  3^  inches 
in  diameter?  Ans.  $5145. 

Questions  Solved  by  the  Rule  of  Three, 
Direct  or  Inverse. 

There  is  a  cistern  which  has  a  stream  of  water 
running  into  it;  it  has  10  cocks;  all  running  to- 
gether will  empty  it  in  2J  hours ;  6  will  empty  it 
in  5 J  hours ;  how  long  will  it  take  3  to  empty  it  ? 

Ans.  55  hours. 

NOTE. — The  6  cocks  will  discharge  in  4£  hours 
what  the  10  cocks  will  in  2J  hours ;  therefore  it 
would  take  the  6  cocks  1 J  to  discharge  what  would 
run  into  the  cistern  in  3  hours :  therefore  it  would 
take  the  6  cocks  1  111  to  discharge  what  would  run 


SQUARE   AND   CUBE   ROOSTS.  141 

in  2^  hours ;  consequently,  2f  cocks  to  discharge 
the  water  as  fast  as  it  run  in. 

There  is  a  stick  of  timher,  12  feet  long,  to  be 
carried  by  3  men :  one  carries  at  the  end,  the  other 
two  carry  by  a  lever ;  how  far  must  the  lever  be 
placed  from  the  other  end  that  each  may  carry 
equally  ?  Ans.  3  feet  from  the  end. 

NOTE. — All  bodies  gravitate  in  an  inverse  pro- 
portion to  the  distance  of  the  center  of  gravity. 

As  1  is  to  6,  the  center,  so  is  2  to  the  answer 
required. 

Man.  Feet.        Men. 

As  1     :     6     :  :    2 
1 

2)6 

3  Ans. 

A  stick  of  timber,  30  feet  long,  to  be  carried  by 
5  men :  two  carry  at  one  end,  the  other  three  l/y  a 
lever ;  how  far  from  the  center  must  the  lever  ta 
placed  that  all  may  carry  equally  ? 

Men.       Feet.        Men. 
As  2    :     15     :  :    3 
2 

3)30 
10  Ans. 


112       ORTON'S  LIGHTNING  CALCULATOR. 
MENSURATION  OR  PRACTICAL  GEOMETM, 

MEASUREMENT   OF   GRINDSTONES. 

Grindstones  are  sold  by  the  stone,  and  their  con* 
tents  found  as  follows:* 

RULE. — To  the  whole  diameter  add  half  of  the 
diameter,  and  multiply  the  sum  of  these  by  the 
same  half,  and  this  product  by  the  thickness ;  di- 
vide this  last  number  by  1728,  and  the  quotient  is 
the  contents ,  or  answer  required. 

EXAMPLES. 

What  are  the  contents  of  a  grindstone  24 
inches  diameter,  and  4  inches  thick 


24+12X24X4 

=1  stone.  An*. 

1728 

2.  What  are  the  contents  of  a  grindstone  36 
inches  diameter,  and  4  inches  thick.    Ans.  2J  stone. 

Mensuration  of  Superficies  and  Solids. 
Superficial  measure  is  that  which  relates  to  length 
and  hreadth  only,  not  regarding  thickness.  It  is 
made  up  of  squares,  either  greater  or  less,  accord  • 
ing  to  the  different  measures  by  which  the  dimen- 
sions of  the  figure  are  taken  or  measured.  Land 
is  measured  by  this  measure,  its  dimensions  being 

*24  inches  in  diameter,  and  4  inches  thick  make  a  stone. 


MENSURATION  OR  PRACTICAL  GE3MET1TY.   14& 

usually  taken  in  acres,  rods,  and  links.  The  con- 
tents of  boards,  also,  are  found  by  this  measure, 
their  dimensions  being  taken  in  feet  and  inches. 
Because  12  inches  in  length  make  1  foot  of  long 
measure,,  therefore  12X12^=144,  the  square  inches 
in  a  superficial  foot,  etc. 
NOTE. — Superficial  means  lying  on  the  surface* 

To  find  the  area  of  a  square  having  equal  sides. 

RULE. — Multiply  the  side  of  the  square  into 
itself,  and  the  product  will  be  the  area,  or  superfi- 
cial content  of  the  same  name  with  the  denomina- 
tion taken,  whether  inches,  feet,  yards,  rods>  and 
links,  or  acres. 

EXAMPLES. 

1.  How  many  square  feet  of  boards  are  contain- 
ed in  the  floor  of  a  room  \vhich  is  20  feet  square  ? 

20X20=400  feet,  the  answer. 

2.  Suppose  a  square  lot  of  land   measures   36 
rods  on  each  side,  how  many  acres  does  it  contain  ? 

36X36=1296  square  rods.     And 
1296-^-160=8  acres,  16  rods,  Ans. 

As  160  square  rods  make  an  acre,  therefore  we  di- 
\ide  1296  by  160  to  reduce  rods  to  acres. 

N.  B. — The  shortest  way  to  work  this  example 
is,  to  cancel  36X36  with  the  divisor  160.  Arrange 
the  example  as  below ;  (divide  both  terms  by  4X^ :) 

36X36  9X9 

same  as =8.1  acres,  or  Sac.  16 

160  10 


144       ORTON'S  LIGHTNING  CALCULATOR. 

To  measure  a  parallelogram  or  long  square. 
RULE.— Multiply  the  length  by  the  breadth,  and 
the  product  will  be  the  area,  or  superficial  content, 
in  the  same  name  as  that  in  which  the  dimension 
teas  taken,  whether  inches,,  feet,  or  rods,  etc. 
EXAMPLES 

1.  A  certain  garden,  in  form  of  a  long  square,  is 
96  feet  long,  and  54  feet  wide;  how  many  square 
feet  of  ground  are  contained  in  it  ? 

Ans.  96X54=5184  square  feet. 

2.  A  lot  of  land,  in  form  of  a  long  square,  is 
120  rods  in  length,  and  60  rods  wide ;  how  many 
acres  are  in  it?     120X60=7200  sq.  rods.     And 
7200^-160=45  acres,  Ans. 

NOTE. — The  learner  must  recollect  that  feet  in 
length,  multipled  by  feet  in  breadth,  produce  square 
feet ;  and  the  same  of  the  other  denominations  of 
lineal  measure. 

NOTE. — Both  the  length  and  breadth,  if  not  in 
units  of  the  same  denomination,  must  be  made  so 
before  multiplying. 

3.  How  many  acres  are  in  a  field  of  oblong 
form,  whose  length  is  14,5  chains,  and  breath  9,75 
chains?  Ans.  14ac.  Orood,  22rods. 

NOTE. — The  Gunter's  chain  is  66  feet,  or  4  rods, 
iOng,  and  contains  100  links.  Therefore  if  dimen- 
sions be  given  in  chains  and  decimals,  point  off 
from  the  product  one  more  decimal  place  than  are 


MENSURATION  OE  PRACTICAL  GlftMETflY.   145 

contained  in  both  factors,  and  it  will  be  acres  and 
decimals  of  an  acre ;  if  in  chains  and  links,  do  the 
Bame,  because  links  are  hundredths  of  chains,  and 
therefore  the  same  as  decimals  of  them.  Or,  as  1 
chain  wide,  and  10  chains  long,  or  10  square  chains, 
or  100000  square  links,  make  an  acre,  it  is  the  same 
as  if  you  divide  the  links  in  the  area  by  100000. 

4.  If  a  board  be  21  feet  long  and  18  inches 
broad,  how  many  square  feet  are  contained  in  it? 

18  inches— 1,5  foot;  and  21x1,5=31,5  ft.,  Ans. 

Or,  in  measuring  boards,  you  may  multiply  the 
length  in  feet  by  the  breadth  in  inches,  and  divide 
the  product  by  12 ;  the  quotient  will  give  the  an- 
swer in  square  feet,  etc.  21x18 

Thus,  in  the  last  example,  =31Jsq.  ft., 

as  before.  12 

5.  If  a  board  be  8  inches  wide,  how  much  :a 
length  will  make  a  foot  square  ? 

BULE. — Divide  144  by  the  width;  thus,  8)144 

Ans.  18  in. 

6.  If  a  piece  of  land  be  5  rods  wide,  how  many 
rods  in  length  will  make  an  acre  ? 

RULE. — Divide  160  by  the  width,  and  the  quo- 
tient will  be  the  length  required;  thus, 
5)160 

Ans.  32  rods  in  length. 


146       ORTON'S  LIGHTNING  CALCULATOR. 

NOTE. — When  a  board,  or  any  other  surface,  is 
wider  at  one  end  than  the  other,  but  yet  is  of  a 
true  taper,  you  inay  take  the  breadth  in  the  middle.. 
or  add  the  widths  of  both  ends  together,  and  halve 
the  sum  for  the  mean  width ;  then  multiply  the 
said  mean  breadth  in  either  case  by  the  length ; 
the  product  is  the  answer  or  area  sought. 

7.  How  many  square  feet  in  a  board,  10  feet 
long  and  13  inches  wide  at  one  end,  and  9  inches 
wide  at  the  other?         13+9 

=11  in.,  mean  width. 

2       ft.     in. 

10x11 

=9Jft.,  Ans. 

12 

8.  How  many  acres  are  in  a  lot  of  land  which  is 
40  rods  long,  and  30  rods  wide  at  one  end,  and  20 
rods  wide  at  the  other  ? 

30+20 

=25  rods,  mean  width. 

2  Then,  25x40 

=6^  acres,  Ans. 

160 

9  If  a  farm  lie  250  rods  on  the  road,  and  at  one 
end  be  75  rods  wide,  and  at  the  other  55  rods  wide, 
bow  many  acres  does  it  contain  ? 

Ans.  101  acres,  2  roods,  10  rods. 

N.  B. — Always  arrange  your  example  as  above 
and  cancel  the  factors  common  to  both  terms  before 
multiplying 


>iJ»tf  PS  RATION  OR  PRACTICAL  GEOMETRY.   147 


3.  —  To  measure  the  surface  of  a  triangle. 

DEFINITION.  —  A  triangle  is  any  three-cornered 
£p:ire  which  is  bounded  by  three  right  lines.* 

RULE-  —  Multiply  the  base  of  the  given  triangle 
mto  half  its  perpendicular  hight,  or  half  the  base 
mto  the  whole  perpendicular  ,  and  the  product  will 

kc  the  area. 

EXAMPLES. 

1.  Required  the  area  of  a  triangle  whose  hase  or 
longest  side  is  32  inches,  and  the  perpendicular 
bight  14  inches. 

14-j-2=7=half  the  perpendicular.     And 
32X7—  224sq.  in.,  Ans. 

2.  There  is  a  triangular  or  three-cornered  lot  of 
land  whose  base  or  longest  side  is  51  J  rods;  the 
perpendicular,  from  the  corner  opposite  to  the  hase, 
measures  44  rods  ;  how  many  acres  does  it  contain  ? 

44-j-2=22=half  the  perpendicular. 
And  51,5X22 

---  =7  acres,  13  rods,  Ans. 
160 

Joists  and  planks  are  measured  l>y  the  following  : 

RULE.  —  Find  the  area  of  one  side  of  the  joist  or 

plank  by  one  of  the  preceding  rules  ;  then  multiply 

it  by  the  thickness  in  inches,  end  the  last  product 

will  be  the  superficial  content. 

*  A  triangle  may  be  either  right-angled  or  oblique. 
10  N 


148          ORTON'S  LIGHTNING  CALCULATOR. 

EXAMPLES. 

1.  What  is  the  area,  or  superficial  content,  ,r 
board  measure,  of  a  joist,  20  feet  long,  4  inches 
wide,  and  3  incLes  thick?     20X4 

X3=20ft.,  Ans. 

12 

2.  If  a  plank  be  32  feet  long,  17  inches  wide, 
and  3  inches  thick,  what  is  the  board  measure  of 
it  ?  Ans.  136  feet 

NOTE. — There  are  some  numbers,  the  sum  of 
whose  squares  makes  a  perfect  square ;  such  are  3 
and  4,  the  sum  of  whose  squares  is  25,  the  square 
root  of  which  is  5 ;  consequently,  when  one  leg 
of  a  right-angled  triangle  is  3,  and  the  other  4, 
the  hypotenuse  must  be  5.  And  if  3,  4,  and  5,  be 
multiplied  by  any  other  numbers,  each  by  the  same, 
the  products  will  be  sides  of  true  right-angled  tri- 
angles. Multiplying  them  by  2,  gives  6,  8,  and  10, 
by  3,  gives  9,  12,  and  15 ;  by  4,  gives  12,  16,  and 
20,  etc.;  all  which  are  sides  of  right-angled  tri- 
angles. Hence  architects,  in  setting  off  the  corners 
of  buildings,  commonly  measure  6  feet  on  one  side, 
and  8  feet  on  the  other ;  then,  laying  a  10-foot  pole 
across  from  those  two  points,  it  makes  the  corner 
a  true  right-angle. 

N.  B. — The  solutions  of  the  foregoing  problems 
are  all  very  brief  by  canceling. 


MENSURATION  OR  PRACTICAL  GEOMETRY.     149 

To  find  tlie  area,  of  any  triangle  when  the  three  sides 

only  are  given. 

RULE. — From  half  the  sum  of  the  three  sides  sub- 
tract each  side  severally ;  multiply  these  three  re- 
mainders and  the  said  half  sum  continually  together  ; 
then  the  square  root  of  the  last  product  will  be  the 
area  of  the  triangle. 

EXAMPLE. 

Suppose  I  have  a  triangular  fish-pond,  whose 
three  sides  measure  400,  348,  and  312yds  j  what 
quantity  of  ground  does  it  cover  ? 

Ans.  10  acres,  3  roods,  8-f  rods. 

NOTE. — If  a  stick  of  timber  be  hewn  three 
square,  and  be  equal  from  end  to  end,  you  find 
the  area  of  the  base,  as  in  the  last  question,  in 
inches ;  multiply  that  area  by  the  whole  length, 
and  divide  the  product  by  144,  to  obtain  the  solid 
content. 

If  a  stick  of  timber  be  hewn  three  square,  be  12 
feet  long,  and  each  side  of  the  base  10  inches,  the 
perpendicular  of  the  base  being  8f  inches,  what  is 
*4s  solidity?  Ans.  3,6+feet. 

PROBLEM  1. 

The  diameter  given,  to  find  the  circumference. 

RULE. — As  1  are  to  22,  so  is  the  given  diameter 
to  the  circumference;  or,  more  exactly,  as  113  are 
to  355,  50  is  the  diameter  to  the  circumference,  etc 


150          ORTON^S   LIGHTNING   CALCULATOR. 

EXAMPLES. 

1.  'What  is  the  circumference  of  a  wheel,  whose 
diameter  is  4  feet  ? 

As  7  :  22  :  :  4  :  12,57+ft,  the  cirram.,  An*. 

2.  What  is  the  circumference  of  a  circle,  whose 
diameter  is  35  rods  ? 

As  7  :  22  :  :  35  :  110  rods,  Am. 

NOTE. — To  find  the  diameter  when  the  circum- 
ference is  given,  reverse  the  foregoing  rule,  and  say, 
as  22  are  to  7,  so  is  the  given  circumference  to  the 
required  diameter;  or,  as  355  are  to  113,  so  is  the 
circumference  to  the  diameter. 

3.  What  is  the  diameter  of  a  circle,  whose  cir- 
cumference is  110  rods? 

A.S  22  :  7  :  :  110  :  35  rods,  the  diam.,    Am. 

CA.SE  5. —  To  find  how  many  solid  feet  a  round 
stick  of  timber,  equally  thick  from  end  to  endy 
will  contain ,  when  hewn  square. 
RULE. — Multiply  twice  the  square  of  its  semi-di- 
ameter, in  inches,  by  the  length  in  the  feet;  then  divide 
the  product  by  144,  and  the  quotient  will  be  the  an- 
swer. 

N.  B. — When  multiplication  and  division  aro 
combined,  always  cancel  like  factors.  When  the 
numbers  are  properly  arranged,  a  lew  clips  with 
the  pencil;  and,  perhaps,  a  irijting  multiplie^on 
will  suffice. 


MENSURATION  OE  PRACTICAL  GEtfME^HY.      151 


EXAMPLES. 

1.  If  the  diameter  of  a  round  stick  of  timber 
be  22  inches,  and  its  length  20  feet,  how  many  solid 
feet  will*  it  contain  when  hewn  square  ? 
11XHX2X20 
Half  dismeter=ll,  and  --  =33,  6+ft. 

144 

the  solidity  when  hewn  square,  the  answer. 
CASE  6.  —  To  find  how  many  feet  of  square  edged 
boards,  of  a  given  thickness,  can  be  sawn  from  a 
log  of  a  given  diameter. 

RULE.  —  Find  the  solid  content  of  the  log,  when 
made  square,  by  the  last  case;  then  say,  as  the 
thickness  of  the  board,  including  the  saw  calf,  is  to 
the  solid  feet,  so  is  12  inches  to  the  number  of  feet 
of  boards. 

EXAMPLES. 

1.  How  many  feet  of  square  edged  boards,  1J 
inch  thick,  including  the  saw  calf,  can  be  sawn  from 
a  log  20  feet  long,  and  24  inches  diameter? 
12X12X2X20 

--  =40ft.  solid  content  when  hewn  sq. 
144 

As  1J  :  40  :  :  12  :  384  feet,  Ans. 

2.  How  many  feet  of  square  edged  boards,  1J 
inch  thick,  including  the  saw  gap,  can  be  sawn  from 
a  log  12  feet  long,  and  18  inches  diameter? 

Ans.  108  feet 


152  ORTON-'S  LIGHTNING  CALCULATOR. 

NOTE. — A  short  rule  for  finding  the  number  of 
feet  of  one  inch  boards  that  a  log  will  make,  is  to 
deduct  J  of  its  diameter  in  inches,  and  J  of  its 
length  in  feet ;  then  for  each  inch  of  diameter  that 
remains,  reckon  1  board  of  the  same  width  as  this 
reduced  diameter,  and  of  the  same  length  as  this 
reduced  length  of  the  log :  thus  a  log  12  feet  long, 
and  12  inches  through,  gives  9  boards,  9  feet  long, 
9  inches  wide,  or  60f  feet — a  log  16  feet  long,  and 
16  inches  through,  gives  12  boards,  12  inches 
wide,  12  feet  long,  or  144  feet. 

In  measuring  timber,  however,  you  may  multiply 
the  breadth  in  inches  by  the  ^epth  in  inches,  and 
that  product  by  the  length  in  feet;  divide  this  last 
product  by  144  and  the  quotient  will  be  the  solid 
content  in  feet,  etc. 

How  many  solid  feet  does  a  piece  of  square  tim- 
ber, or  a  block  of  marble    contain,  if   it  be  16 
inches  broad,  11  inches  thick,  and  20  feet  long? 
16X 11 X 20=3520,  and  35  2  0--1 44=24,4+ sol.  ft. 
CASE  8. —  To  Jlnd  the  solidity  of  a  cone  or  pyramid 

whether  round,  square,  or  triangular. 

DEFINITION. — Solids  which  decrease  gradually 
from  the  base  till  they  come  to  a  point,  are  generally 
called  cones  or  pyramids,  and  are  of  various  kinds, 
according  to  the  figure  of  their  bases ;  round, 
square,  oblong,  triangular,  etc.;  the  point  at  the 
top  is  called  the  vertex,  and  a  line  drawn  frcrn  the 


MENSURATION  OR  PRACTICAL  GEOMETIJJ.   153 

vertex,  perpendicular  to  the  base,  is  called  the 
flight  of  the  pyramid. 

RULE. — Find  the  area  of  the  base,  whether  roundt 
square,  oblong,  or  triangularly  some  one  oj  the  fort 
going  rules,   as  the  case  may  be;  then  multiply  this 
area  by  one-third  of  the  hight,  and  the -product  will 
be  the  solid  content  of  the  pyramid. 

EXAMPLES. 

1.  What  is  the  content  of  a  true-tapered  round 
stick  of  timber,  24  feet  perpendicular  length,  15 
inches  diameter  at  one  end,  and  a  point  at  the  other  ? 
15xl5x,7854x8 

'• =9,8175  solid  feet,  Ans. 

144 

To  find  the  solid  content  of  a  frustum  af  a  cone. 
What  is  the  solid  content  of  a  tapering  round 
stick  of  timber,  whose    greatest   diameter  is  13 
inches,  the  least  6J  inches,  and  whose  length  is  24 
feet,  calculating  it  by  both  rules  ? 

RULE.  2  — Multiply  each  diameter  into  itself;  mul- 
tiply one  diameter  by  the  other;  multiply  the  sum 
of  these  products  by  the  lengths  ;  annex  two  ciphers 
to  the  product,  and  divide  itr  by  382 ;  the  quotient 
will  be  the  content ,  which  divide  by  144  for  feet  a* 
in  other  cases. 
;13X13)+(6,5X6,5)+(13X6,5)X2400 

=1858,115+ 

382 
And  1858,115-j-144=12,903-i-ft,  Am. 


154       ORTON'S  LIGHTNING*  CALCULATOR. 

To  find  the  content  of  timber  in  a  tree,  multiply 
the  square  of  i  of  the  circumference  at  the  middle 
of  the  tree,  in  inches,  by  twice  the  length  in  feet, 
and  the  product  divided  by  144  will  be  the  content, 
extremely  near  the  truth.  In  oak,  an  allowance 
of  ye  or  T\j  must  be  made  for  the  bark,  if  on 
the  tree;  in  other  wood,  less  trees  of  irregular 
growth,  must  be  measured  in  parts. 

To  find  the  solid  content  of  a  frustum  or  segment 
of  a  globe. 

DEFINITION. — The  frustum  of  a  globe  is  any  part 
cut  off  by  a  plane. 

RULE. —  To  three  times  the  square  of  the  semi-di- 
ameter of  the  base,  add  the  square  of  the  hight; 
multiply  this  sum  by  the  hight,  and  the  product  again 
by  .5236 ;  the  last  product  will  be  the  solid  content. 

EXAMPLE. 

If  the  hight  of  a  coal-pit,  at  the  chimney,  be  9 
feet,  and  the  diameter  at  the  bottom  be  24  feet, 
how  many  cords  of  wood  does  it  contain,  allowing 
nothing  for  the  chimney? 
24-=-2=rl2=:h'fdiain.  12X12X3=432.    9X9=81 

And  432+ 81 X9X, 5236 

18,886+ cords,  Am. 

128=rsolid  feet  in  a  cord. 


MENSURATION  OR  PRACTICAL  aEOJgSTRjjT.    155 


NOTE.  —  A  pile  of  wood  that  is  8  feet  long,  4  feet 
high,  and  4  feet  wide,  contains  128  cubic  feet,  or 
a  cord,  and  every  cord  contains  8  cord-feet;  and 
as  8  is  y1^  -of  128,  every  cord-foot  contains  16  cubic 
feet;  therefore,  dividing  the  cubic  feet  in  a  pile  of 
wood  by  16,  the  quotient  is  the  cord-feet;  and  if 
cord-feet  be  divided  by  8,  the  quotient  is  cords. 

NOTE.  —  If  we  wish  to  find  the  circumference  of 
a  tree,  which  will  hew  any  given  number  of  inches 
square,  we  divide  the  given  side  of  the  square  by 
.225,  and  the  quotient  is  the  circumference  re- 
quired. 

What  must  be  the  circumference  of  a  tree  that 
will  make  a  beam  10  inches  square? 

NOTE.  —  When  wood  is  "  corded"  in  a  pile  4  feet 
wide,  by  multiplying  its  length  by  its  hight,  and 
dividing  the  product  by  4,  the  quotient  is  the  cord- 
feet  ;  and  if  a  load  of  wood  be  8  feet  long,  and  its 
hight  be  multiplied  by  its  width,  and  the  product; 
divided  by  2,  the  quotient  is  the  cord-feet. 

How  many  cords  of  wood  in  a  pile  4  feet  wide, 
70  feet  6  inches  long,  and  5  feet  3  inches  high? 

NOTE.  —  Small  fractions  rejected. 

To  find  how  large  a  cube  may  be  cut  from  any 
given  sphere,  or  be  inscribed  in  it. 

RULE.  —  Square  the  diameter  of  the  sphere,  divide 
that  product  ly  3,  and  extract  the  square  root  of  the 
quotient  for  the  answer. 


116        ORTON'S  LIGHTNING  CALCULATOR. 

To  find  the  contents  of  a  round  vessel,  widei  at 
one  end  than  the  other. 

RULE. — Multiply  the  great  diameter  by  the  less, 
to  this  product  add  \  of  the  square  of  their  differ* 
ence,  then  multiply  by  the  hight,  and  divide  as  in 
the  last  rule. 

Having  the  diameter  of  a  circle  given,  to  find 
the  area. 

RULE. — Multiply  half  the  diameter  by  half  the 
circumference,  and  the  product  is  the  area;  or, 
which  is  the  same  thing,  multiply  the  square  of  the 
diameter  by  .7854,  and  the  product  is  the  area. 

To  find  the  solidity  of  a  sphere  or  globe. 

RULE. — Multiply  the  cube  of  the  diameter  by 
.5236. 

To  find  the  convex  surface  of  a  sphere  or  globe. 

RULE. — Multiply  its  diameter  by  its  circumfer- 
ence. 

To  find  the  solidity  of  a  prism. 

RULE — Multiply  the  area  of  the  base,  or  end,  by 
the  hight. 

How  many  wine  gallons  will  a  cubical  box  con- 
tain, that  is  10  long,  5  feet  wide,  and  4  feet  high? 

RULE. —  Take  the  dimensions  in  inches;  then  mul- 
tiply the  length,  breadth,  and  hight  together;  di- 
vide the  product  by  282  for  ale  gallons,  231  for  icine 
and  2150  for  bushels. 


MENSURATION  OR  PRACTICAL  GEOMETRY.     157 

I  have  a  piece  of  timber,  30  inches  in  diameter ; 
how  large  a  square  stick  can  be  hewn  from  it  ? 

RULE. — Multiply  the  diameter  by  .7071,  and  the 
produot  is  the  side  of  a  square  inscribed. 

I  have  a  circular  field,  360  rods  in  circumference ; 
what  must  be  the  side  of  a  square  field  that  shall 
contain  the  same  quantity? 

RULE. — Multiply  the  circumference  by  .282,  and 
the  product  is  the  side  of  an  equal  square. 

I  have  a  round  field,  50  rods  in  diameter;  what 
is  the  side  of  a  square  field  that  shall  contain  the 
same  area?  Ans.  44.31135-|-rods. 

RULE. — Multiply  the  diameter  by  .886,  and  the 
product  is  the  side  of  an  equal  square. 

There  is  a  certain  piece  of  round  timber,  30 

inches  in  diameter ;  required  the  side  of  an  equi- 

.  lateral  triangular  beam  that  may  be  hewn  from  it. 

RULE. — Multiply  the  diameter  by  .866,  and  the 
product  is  the  side  of  an  inscribed  equilateral  tri- 
angle. 

To  find  the  area  of  a  globe  or  sphere. 

DEFINITION. — A  sphere  or  globe  is  a  round  solid 
body,  in  the  middle  or  center  of  which  is  an  imag- 
inary point,  from  which  every  part  of  the  surface 
is  equally  distant.  An  apple,  or  a  ball  used  by 
children  in  some  of  their  pastimes>  may  be  called 
a  sphere  or  globe. 


158       ORTON'S  LIGHTNING  CALCULATOR. 

RULE. — Multiply  the  circumference  by  the  diam- 
eter, and  the  product  will  be  tlie  area  or  surface. 

To  find  the  force  of  the  wedge. 

RULE. — As  half  the  breadth  or  thicJtness  of  the 
head  of  the  wedge  is  to  one  of  its  slanting  sides,  so  is 
the  power  which  acts  against  its  head  to  the  force 
produced  at  its  side. 

Suppose  100  pounds  to  be  applied  to  the  head 
of  a  wedge  that  is  2  inches  broad,  and  whose  slant 
is  20  inches  long,  what  force  would  be  affected  on 
each  side?  * 

To  find  the  solidity  of  a  cone  or  pyramid. 

RULE. — Multiply  the  area  of  the  base  by  one-third 
of  its  height,  or  vice  versa. 

II.   TIMBER   MEASUBE 

PROBLEM  1. 
To  find  the  superficial  contents  of  a  board  or  plank  > 

RULE. — Multiply  the  length  by  the  breadth. 

NOTE. — When  the  board  is  broader  at  one  end 
than  the  other,  add  the  breadth  of  the  two  ends 
together,  and  take  half  the  sum  for  a  mean 
breadth. 

N.  B. — When  the  breadth  of  the  board  is  in 
inches,  or  feet  and  inches : 

RULE. — Multiply  the  length  of  the  board,  taken  in 
feet,  by  its  breadth  taken  in  inches,  and  divide  this 
produci  by  1 2 ;  the  quotient  is  the  contents  in  square 


TIMBER   MEASURE.  159 

PROBLEM  III. 
To  find  tlit  solid  contents  of  squared  or  fow+iided 

Timber. 

By  the  Carpenters1  Rule. 

As  12  on  D  :  length  on  c  :  Quarter  girt  on  D  : 
solidity  on  c. 

RULE  I. — Multiply  the  breadth  in  the  middle  by 
the  depth  in  the  middle,  and  that  product  by  the 
length  for  the  solidity. 

NOTE. — If  the  tree  taper  regularly  from  one  end 
to  the  other,  half  the  sum  of  the  breadths  of  the 
two  ends  will  be  the  breadth  in  the  middle,  and 
half  the  sum  of  the  depths  of  the  two  ends  will  bo 
the  depth  in  the  middle. 

RULE  II. — Multiply  the  sum  of  the  breadths  of 
the  two  ends  by  the  sum  of  the  depths,  to  which  add 
the  product  of  the  breadth  and  depth  of  each  end  ; 
one-sixth  of  this  sum  multiplied  by  the  length,  wilt 
give  the  correct  solidity  of  any  piece  of  squared  tim- 
ber tapering  regularly. 

PROBLEM  IV. 

Ft?  find  how  much  in  length  will  make  a  solid  foot, 

or  any  other  asssigned  quantity,  of  squared  timber, 

of  equal  dimensions  from  end  to  end. 

RULE. — Divide  1728,  the  solid  inches  in  a  foot 

+r  the  solidity  to  be  cut  off,  by  the  area  of  the  end 

tw  inches,  and  the  quotient  will  be  the  length  in  inches. 


160       ORTON'S  LIGHTNING  CALCULATOR. 

NOTE. — To  answer  the  purpose  of  the  above 
rule,  some  carpenters'  rules  have  a  little  table  upon 
them,  in  the  following  form,  called  a  table  of  tim- 
ber measure. 


0    |   0   |   0   |  0  |  9  |  0  |  11     3     9     inches. 


I  144  |  36  |  16  |  9  |  5  |  4  |   2   |  2  |  1  |  feet. 


1    |2|3|4|5|6|7|8|9|  side  of  the  square. 


This  table  shows,  that  if  the  side  of  the  square 
be  1  inch,  the  length  must  be  144  feet ;  if  2  inches 
be  the  side  of  the  square,  the  length  must  be  36 
feet,  to  make  a  solid  foot. 

PROBLEM  V. 
To  find  the  solidity  of  round  or  unsquared  timber 

RULE  I. —  Gird  the  timber  round  the  middle  with 
a  string ;  one-fourth  part  of  this  girt  squared  and 
multiplied  by  the  length  will  give  the  solidity. 

NOTE. — If  the  circumference  be  taken  in  inches, 
and  the  length  in  feet,  divide  the  last  product  by 
144. 

RULE  II — By  the  Table. — Multiply  the  area  cor- 
responding to  the  quarter-girt  in  inches,  by  the 
length  of  the  piece  of  timber  in  feet,  and  the  product 
will  be  the  solidity. 

NOTE. — If  the  quarter  girt  exceed  the  table,  take 
half  of  it,  and  four  times  the  content  thus  formed 
will  be  the  answer. 


TIMBER   MEASURE. 


161 


A  TABLE  FOR  MEASURING  TIMBER. 


Quarter 
Girt. 

Area. 

Quarter 
Girt. 

Area. 

Quarter 
Girt. 

Area. 

Inches. 

6 
6} 
6| 

6J 

"  Feet. 

.250 
.272 

.294 
.317 

Inches. 

12 
12J 
12} 

12f 

Feet. 

1.000 
1.042 
1.085 
1.129 

Indies. 

18 
18} 
19 
19} 

Feet. 
2.250 

2.376 
2.506 
2.640 

7 
7J 
7J 
7J 

.340 
.364 
.390 
.417 

13 
13} 
13} 

13| 

1.174 
1.219 
1.265 
1.313 

20 
20} 
21 
21} 

2.777 
2.917 
3.062 
3.209 

i-to  r-ikM  «|r)< 
OO  OO  00  00 

.444 
.472 
.501 
.531 

14 
14t 
14} 

14f 

1.361 
1.410 
1.460 
1.511 

22 

22} 
23 
23} 

3.362 
3.516 
3.673 
3.835 

9 
9J 
91 

9J 

.562 
.594 
.626 
.659 

15 
15} 
15} 
15| 

1.562 
1.615 
1.668 
1.722 

24 
24} 

25 

25} 

4.000 
4.168 
4'.  340 
4.516 

10 

iot 
io| 

10} 

.694 
.730 
.766 
.803 

16 
16} 
16} 
16| 

1.777 
1.833 

1.890 
1.948 

26 

26} 
27 
27} 

4.694 
4.876 
5.062 
5.252 

11 

Hi 

11} 

ll! 

.840 
.878 
.918 
.959 

17 

m 

17} 

17| 

2.006 
2.066 
2.126 

2.187 

28 
28} 
29 
29} 
30 

5.444 
5.640 

5.840 
6.044 
6.250 

RULE  III — By  the  Carpenters  Rule. — Measure 
Ihe  circumference  of  the  piece  of  timber  in  the  middle 
wnd  take  a  •  ??  trier  of  it  in  inches,  call  this  the  girt. 


162 


ORTON  S   LIGHTNING   CALCULATOR. 


Then  set  12  on  i>,  to  the  length  in  feet  on  c,  and 
against  the  girt  in  inches  on  D,  you  will  find  the 
content  in  feet  on  c. 

EXAMPLE  1. 

If  a  piece  of  round  timber  be  18  feet  long,  and 
the  quarter  girt  24  inches,  how  many  feet  of  timbei 
are  contained  therein  ? 

24  quarter  girt. 
24 


96 

48 

576  square. 
18 


4608 
576 


By  the  Table. 

Against  24  stands    4.00 
Length,        18 


Product, 

Ans.  72  feet. 


72.00 


144)10368(72  feet 
1008 

288 
288 

By  the  Carpenters'  Rule. 
12  on  D  :  18  on  c  :  24  on  D  :  72  on  c. 

in.    CARPENTERS'  AND  JOINERS'  WORK. 
The  Carpenters'  and  Joiners'  works,  which  are 
measurable,    are     flooring,    partitioning,    roofing 
wainscoting,  etc. 


TIMBER  MEASURE. 


163 


1.   Of  Flooring. 

Joists  are  measured  by  multiplying  their  breadth 
by  their  depth,  and  that  product  by  their  length. 
They  receive  various  names,  according  to  the  posi- 
tion in  which  they  are  laid  to  form  a  floor,  such  as 
trimming  joists,  common  joists,  girders,  binding 
joists,  bridging  joists  and  ceiling  joists. 

Girders  and  joists  of  floors,  designed  to  bear 
great  weights,  should  be  let  into  the  walls  at  each 
end  about  two-thirds  of  the  wall's  thickness. 

In  boarded  flooring,  the  dimensions  must  be 
taken  to  the  extreme  parts,  and  the  number  of 
squares  of  100  feet  must  be  calculated  from  these 
dimensions.  Deductions  must  be  made  for  stair- 
cases, chimneys,  etc. 

Example  1.  If  a  floor  be  57  feet  3  inches  long, 
and  28  feet  6  inches  broad,  how  many  squares  of 
flooring  are  there  in  that  room  ? 


By  Decimals. 
57.25 

28.5 

By  Duodecimals. 
F.         I. 
57    :    3 

28    :    6 

28625 
45800 
11450 

456 
114 

28 

7 

:    7    :    6 
:    0    :    0 

100)1631.625  feet. 

Squares  16.31625 


11 


16:31   :    7    :    6 
Ans.  16  squares  and  31  feet, 


164         ORTON'S  LIGHTNING  CALCULATOR. 

iv.     OF  BRICKLAYERS'  WORK. 

The  principal  is  tiling,  slating,  walling  and  chha- 
aey  work. 

1.    Of  Tiling  or  Slating. 

Tiling  and  slating  are  measured  by  the  square 
of  100  feet,  as  flooring,  partitioning  and  roofing 
were  in  the  Carpenters'  work ;  so  that  there  is  not 
much  difference  between  the  roofing  and  tiling; 
yet  the  tiling  will  be  the  most ;  for  the  bricklayers 
sometimes  will  require  to  have  double  measure  for 
hips  and  valleys. 

When  gutters  are  allowed  double  measure,  the 
way  is  to  measure  the  length  along  the  ridge-tile, 
and  add  it  to  the  content  of  the  roof:  this  makes 
an  allowance  of  one  foot  in  breadth,  the  whole 
length  of  the  hips  or  valleys.  It  is  usual  also  to 
allow  double  measure  at  the  eaves,  so  much  as  the 
projector  is  over  the  plate,  which  is  commonly 
about  18  or  20  inches. 

Sky-lights  and  chimney  shafts  are  generally  de- 
ducted, if  they  be  large,  otherwise  not. 

Example  1.  There  is  a  roof  covered  with  tiles, 
whose  depth  on  both  sides  (with  the  usual  allow- 
ance at  the  eaves)  is  37  feet  3  inches,  and  the 
length  45  feet;  how  many  squares  of  tiling  are 
contained  therein  ? 


BRICKLAYERS'  WORK,  165 


BT    DUODECIMALS. 

FEET.    INCHES. 

37     3 
45     0 

185 
148 
11     3 


BY   DECIMALS. 

37.25 
45 


18625 
14900 


16  76.25 


16  76     3 

2.   Of  Walling. 

Bricklayers  commonly  measure  their  work  by 
the  rod  of  16J  feet,  or  272J  square  feet.  In  some 
places  it  is  a  custom  to  allow  18  feet  to  the  rod  ; 
that  is,  324  square  feet.  Sometimes  the  work  is 
measured  by  the  rod  of  21  feet  long  and  3  feet 
high,  that  is,  63  square  feet ;  and  then  no  regard 
is  paid  to  the  thickness  of  the  wall  in  measuring! 
but  the  price  is  regulated  according  to  the  thick- 
ness. 

When  you  measure  a  piece  of  brick-work,  the 
.first  thing  is  to  inquire  by  which  of  these  ways  it 
must  be  measured ;  then,  having  multiplied  the 
length  and  breadth  in  feet  together,  divide  the  pro- 
duct by  the  proper  divisor,  viz.:  272.25,  324  or  fi3, 
according  to  the  measure  of  the  rod,  and  the  quo- 
tient will  be  the  answer  in  square  rods  cf  that 
measure. 

But,  commonly,  brick  walls  that  are  measured 
by  the  rod  are  to  be  reduced  to  a  standard  thick- 


166         ORTON'S  LIGHTING  CALCULATOR. 

ness  of  a  brick  and  a -half,  which  may  be  done  by 
the  following 

RULE. — Multiply  the  number  of  superficial  feet 
that  are  contained  in  the  wall  by  the  number  of 
half  bricks  which  that  wall  is  in  thickness;  one- 
third  part  of  that  product  will  be  tJie  content  in 
feet. 

The  dimensions  of  a  building  are  generally 
taken  by  measuring  half  round  the  outside  and 
half  round  the  inside,  for  the  whole  length  of  the 
wall ;  this  length,  being  multiplied  by  the  hight, 
gives  the  superficies.  And  to  reduce  it  to  the 
standard  thickness,  etc.,  proceed  as  above.  All  the 
vacuities,  such  as  doors,  windows,  window  backs, 
etc.,  must  be  deducted. 

To  measure  any  arched  way,  arched  window  or 
door,  etc.,  take  the  hight  of  the  window  or  dooi 
from  the  crown  or  middle  of  the  arch  to  the  bot- 
tom or  sill,  and  likewise  from  the  bottom  or  sill  to 
the  spring  of  the  arch ;  that  is,  where  the  arch 
begins  to  turn.  Then  to  the  latter  hight  add  twice 
the  former,  and  multiply  the  sum  by  the  width  of 
the  window,  door,  etc.,  and  one-third  of  the  pro- 
duct will  be  the  area,  sufficiently  near  for  practice. 

Example  1.  If  a  wall  be  72  feet  6  inches  long, 
and  19  feet  3  inches  high,  and  5J  bricks  thick, 
how  many  rods  of  brick  work  are  contained  therein, 
when  reduced  to  the  standard  ? 


GLAZIERS    WORK. 

Vii.  GLAZIERS'  WORK. 

Glaziers  take  their  dimensions  in  feet,  inches 
and  eights  or  tenths,  or  else  in  feet  and  hundredth 
parts  of  a  foot,  and  estimate  their  work  by  the 
square  foot. 

Windows  are  sometimes  measured  by  taking  the 
dimensions  of  one  pane,  and  multiplying  its  super- 
ficies by  the  number  of  panes.  But,  more  gen- 
erally, they  measure  the  length  and  breadth  of  the 
window  over  all  the  panes  and  their  frames  for  the 
length  and  breadth  of  the  glazing. 

Circular  or  oval  windows,  as  fan  lights,  etc.,  are 
measured  as  if  they  were  square,  taking  for  their 
dimensions  the  greatest  length  and  breadth,  as  a 
compensation  for  the  waste  of  glass  and  labor  in 
cutting  it  to  the  necessary  forms. 

Example  1.  If  a  pane  of  glass  be  4  feet  8| 
inches  long,  and  1  foot  4J  inches  broad,  how  many 
feet  of  glass  are  in  that  pane  ? 


BY  DUODECIMALS. 

BY  DECIMALS. 

FT. 

IN. 

p. 

4.729 

4 

8 

9 

1.354 

1 

4 

3 

18916 

4 

8 

9 

23645 

1 

6 

11 

0 

14187 

1 

2 

2     3 

4729 

6 

4 

10 

2     3 

6.403066 

Ans.  6  feet  4  inches. 


168 


ORTON  S   LIGHTNING   CALCULATOR. 


VIII.      PLUMBERS    WORK. 


Plumbers'  work  is  generally  rated  at  so  much 
per  pound,  or  by  the  hundred  weight  of  112 
pounds,  and  the  price  is  regulated  according  to  the 
value  of  lead  at  the  time  when  the  work  is  per- 
formed. 

Sheet  lead,  used  in  roofing,  guttering,  etc., 
weighs  from  6  to  12  pounds  per  square  foot,  ac  • 
cording  to  the  thickness,  and  leaden  pipe  varies  in 
weight  per  yard,  according  to  the  diameter  of  its 
bore  in  inches. 

The  following  table  shows  the  weight  of  a  square 
foot  of  sheet  lead,  according  to  its  thickness,  reck- 
oned in  parts  of  an  inch,  and  the  common  weight 
of  a  yard  of  leaden  pipe  corresponding  to  the 
diameter  of  its  bore  in  inches: 


Thickness 
of  Lead. 

Pounds  to  a 
Square  foot. 

Bore  of 
Leaden  Pipe. 

Pounds 
per  yard. 

ft 

5.899 

1 

10 

i 

6.554 

i 

12 

i 

7.373 

ii 

16 

4 

8.427 

H 

18 

i 

9.831 

if 

21 

* 

11.797 

2 

24           I 

MASON  S  WORK. 


169 


Example  1.  A  piece  of  sheet  lead  measures  10 
feet  9  inches  in  length,  and  6  feet  6  inches  ia 
breadth ;  what  is  its  weight  at  8^  pounds  10  a 
square  foot? 


BY  DUODECIMALS 

BY  DECIMALS 

FEET.  INCHES. 

16         9 
6         6 

100         6 
8        4 

6 

FEET. 

16.75 
6.5 

8375 
10050 

108       10 

6 

108.875  feet. 

Then  1  foot  :  8J  pounds  :  :  108.875  feet  I 
898.21875  pounds=8  cwt.  2J  pounds  nearly. 

ix.    MASON'S  WORK 

Masons  measure  their  work  sometimes  hy  the 
foot  solid,  sometimes  hy  the  foot  superficial,  and 
sometimes  by  the  foot  in  length.  In  taking 
dimensions  they  girt  all  their  moldings  as 
joiners  do. 

The  solids  consist  of  blocks  of  marble,  stone 
pillars,  columns,  etc.  The  superficies  are  payo- 
inents,  slabs,  chimney-pieces,  etc. 


170       OBTON'S  LIGHTNING  CALCULATOR. 

V.    PLASTERERS     WORK. 

Plasterers'  work  is  principally  of  two  kinds ; 
namely,  plastering  upon  laths,  called  ceiling,  and 
plastering  upon  walls  or  partitions  made  of  framed 
timber,  called  rendering. 

In  plastering  upon  walls,  no  deductions  are  made 
except  for  doors  and  windows,  because  cornice, 
festoons,  enriched  moldings,  etc.,  are  put  on  after 
the  room  is  plastered. 

In  plastering  timber  partitions,  in  large  ware- 
houses, etc.,  where  several  of  the  braces  and  larger 
timbers  project  from  the  plastering,  a  fifth  part  is 
commonly  deducted.  Plastering  between  their 
timbers  is  generally  called  rendering  between 
quarters. 

Whitening  and  coloring  are  measured  in  tho 
Bame  manner  as  plastering ;  and  in  timbered  par- 
titions, one-fourth,  or  one-fifth  of  the  whole  area  is 
commonly  added,  for  the  trouble  of  coloring  the 
sides  of  the  quarters  and  braces. 

Plasterers'  work  is  measured  by  the  yard  square, 
consisting  of  nine  square  feet.  In  arches,  the  girt 
round  them,  multiplied  by  the  length,  will  give  the 
superficies. 

Example  1. — If  a  ceiling  be  59  feet  6  inches 
iong,  and  24  feet  6  inches  broad ;  how  many  yards 
does  that  ceiling  contain  ? 


CISTERNS.  171 

PROBLEM  L 

To  find  the  solid  content  of  a  Dome,  having  the 
highi  and  the  dimensions  of  its  base  given. 

RULE. — Multiply  the  area  of  the  base  by  the 
Tiight,  and  f  of  the  product  will  be  the  solidity. 

Example  1. — What  is  the  solidity  of  a  dome,  in 
the  form  of  a  hemisphere,  the  diameter  of  the  cir- 
cular base  being  60  feet  ? 

60'X -7854=1:2827.44  area  of  the  base. 

Then  f  (2827.44X30>=:56548.8  cubic  feet, 
Ans. 

[PROBLEM  II. 

To  find  the  superficies  of  a  dome,  having  the  highi 
and  dimensions  of  its  base  given. 

RULE. — Multiply  the  area  of  the  base  by  2,  and 
ihe  product  will  be  the  superficial  content  required  ; 
or,  multiply  the  square  of  the  diameter  of  the  base 
by  1.5708. 

FOR  AN  ELLIPTICAL  DOME. — Multiply  the  two 
diameters  of  the  base  together,  and  that  product  by 
1.5708,  the  last  product  will  be  the  area,  sufficiently 
correct  for  practical  purposes. 

xi.  CISTERNS.     . 

Cisterns  are  large  reservoirs  constructed  to  hold 
water,  and  to  be  permanent,  should  be  made  either 
of  brick  or  masonry. 


172       ORTON'S  LIGHTNING  CALCULATOR. 

It  frequently  occurs  that  they  are  to  be  so  con- 
structed as  to  hold  given  quantities  of  water,  and 
it  then  becomes  a  useful  aad  practical  problem  tc 
calculate  their  exact  dimensions. 

How  do  you  find  the  number  of  hogsheads 
which  a  cistern  of  given  dimensions  will  contain? 

1st.  Find  the  solid  content  of  the  cistern  in 
cubic  inches. 

2d.  Divide  the  content  so  found  by  14553,  and 
the  quotient  will  be  the  number  of  hogsheads. 

If  the  hight  of  a  cistern  be  given,  how  do  you 
find  the  diameter,  so  that  the  cistern  shall  con- 
tain a  given  number  of  hogsheads  ? 

1st.  Reduce  the  hight  of  the  cistern  to  inches, 
arid  the  content  to  cubic  inches. 

2d.  Multiply  the  hight  by  the  decimal  .7854. 

2.  Divide  the  content  by  the  last  result,  and 
extract  the  square  root  of  the  quotient,  which  wiii 
be  the  diameter  of  the  cistern  in  inches. 

EXAMPLE. 

If  the  diameter  of  a  cistern  be  given,  how  do  you 
find  the  hight,  so  that  the  cistern  shall  contain  a 
given  number  of  hogsheads  ? 

1st,  Reduce  the  content  to  cubic  inches. 

2d.  Reduce  the  diameter  to  inches,  and  then  mul- 
tiply its  square  by  the  decimal  .7854. 


BINS  FOB  GRAIN.  17£ 

3d.  Divide  the  content  by  the  last  result,  and 
Sne  quotient  will  be  the  hight  in  inches. 

XII.    BINS  FOR  GRAIN. 

Having  any  number  of  bushels,  how  then  will 
/•ou  find  the  corresponding  number  of  cubic  feet  ? 

Increase  "the  number  of  bushels  one -fourth 
itself,  and  the  result  will  be  the  number  of  cubic 
feet. 

How  will  you  find  the  number  of  bushels  which 
a  bin  of  a  given  size  will  hold  ? 

Find  the  content  of  the  bin  in  cubic  feet ;  then 
diminish  the  content  by  one-fifth,  and  the  resul* 
will  be  the  content  in  bushels. 

How  will  you  find  the  dimensions  of  a  bin  which 
tshall  contain  a  given  number  of  bushels  ? 

Increase  the  number  of  bushels  one -fourth 
itself,  and  the  result  will  show  the  number  of  cubic 
feet  which  the  bin  will  contain.  Then,  when  two 
dimensions  of  the  bin  are  known,  divide  fhe  last 
result  by  their  product,  and  the  quotient  will  be  the 
)ther  dimension. 

A  Log  Table. — Showing  the  number  of  feet  of 
/oards  any  log  will  make  whose  diameter  is  from 
16  to  36  inches  at  the  smallest  end,  and  from  10 
fco  15  feet  in  length. 


1 74        ORTON'S  LIGHTNING  CALCULATOR 


Diametel 

10  Feet 

Diametel 

^ 

Diametel 

12  Feet 

Diami'toi 

13  Feet 

Diameter 

14  Feet 

Diaiiirtei 

15  Feet 

3 

13 

p 

3 

a 

3 

M 

0* 

p 

5T 

p 

5* 

3 

3 

1  3 

CT? 

o 

n 

3 

<£ 

3 

orq 

B 

0 

en 

3 

0 

a 

M 

of 

• 

1 

• 

er 

i 

er 

Cf 

tr 

15 

90 

15 

99 

15 

108 

15 

117 

15 

126 

15 

135 

16 

100 

16 

110 

16 

120 

16 

130 

16 

140 

16 

150 

17 

125 

17 

137 

17 

150 

17 

160 

17 

175 

17 

187 

18 

155 

18 

170 

18 

186 

18 

201 

18 

216 

18 

232 

19 

165 

19 

176 

19 

198 

19 

214 

19 

230 

19 

247 

20 

172 

20 

189 

20 

206 

20 

263 

20 

246 

20 

258 

21 

184 

21 

202 

21 

220 

21 

238 

21 

256 

21 

276 

22 

194 

22 

212 

22 

232 

22 

263 

22 

294 

22 

291 

23 

219 

23 

240 

23 

278 

23 

315 

23 

332 

23 

333 

24 

250 

24 

276 

24 

300 

24 

325 

24 

350 

24 

375 

25 

280 

25 

308 

25 

336 

25 

364 

25 

392 

25 

420 

26 

299 

26 

323 

26 

346 

26 

375 

26 

404 

26 

448 

27 

327 

27 

3u7 

27 

392 

27 

425 

27 

457 

27 

490 

28 

360 

28 

396 

28 

432 

28 

462 

28 

504 

28 

540 

[  29 

376 

29 

414 

29 

451 

29 

.488 

29 

526 

29 

504 

30 

412 

30 

452 

30 

494 

30 

535 

30 

576 

30 

618 

31 

"428 

31 

471 

31 

513 

31 

558 

31 

602 

31 

642 

32 

451 

32 

496 

32 

541 

32 

587 

32 

631 

32 

676 

33 

490 

33 

539 

33 

588 

33 

637 

33 

686 

33 

735 

34 

532 

34 

585 

|34 

638 

34 

691 

34 

T44 

34 

798 

35 

582 

35 

640 

35 

698 

35 

752 

35 

805 

45 

863 

36 

593 

36 

657 

36 

717 

36 

821 

36 

836 

36 

889 

TABLE  FOR  BANKING  AND  EQUATION. 

Showing  the  number  of  days  from  any  date  in 
one  month  to  the  same  date  in  any  other  month. 

Example.  How  many  days  from  the  2d  of  Feb- 
luary  to  the  2d  of  August?  Look  for  February  at 
the  left  hand,  and  August  at  the  top — in  the  angle 
is  181.  In  leap  year,  add  one  day  if  February  be 
included. 


WEIGHTS  AND  MEASURES. 


175 


From 
To 

4 

ft 

o> 
PH 

SH' 

c3 

& 

*C 

ft 

<! 

£, 

3 
& 

<D 

d 

3 
HS 

j>» 
"5 

>-a 

bb 

53 
< 

+3 

PH 

£ 

"d 
O 

> 
o 

K 

8 

A 

Jan  

365 
334 
306 
275 
245 
214 
184 
153 
122 
92 
61 
31 

31 
365 
337 
306 
276 
245 
215 
184 
153 
123 
92 
62 

59 
28 
365 
334 
304 
273 
243 
212 
181 
151 
120 
90 

90 
59 
31 
365 
335 
304 
274 
243 
212 
182 
151 
121 

120 
89 
61 
30 
365 
334 
304 
273 
242 
212 
181 
151 

151 
120 
92 
61 
31 
365 
335 
304 
273 
243 
212 
182 

181 
150 
122 
91 
61 
30 
365 
334 
303 
273 
242 
212 

212 
181 
153 
122 
92 
61 
31 
365 
334 
304 
273 
243 

243 
212 
184 
153 
123 
92 
62 
31 
365 
335 
304 
274 

273 
242 
214 
183 
153 
122 
92 
61 
30 
365 
334 
304 

304 
273 
245 
214 
184 
153 
123 
92 
61 
31 
365 
335 

331 
303 
275 
244 
214 
183 
153 
122 
91 
61 
30 
365 

Feb  

March  
April  

May  

June  

July... 

Sept  

Oct  

Nov  

Dec  

TABLE    SHOWING    DIFFERENCE  OF   TIME   AT    1? 
O'CLOCK  (NOON) 

New  York.........  12.00  N. 

Buffalo 11.40  A.  M. 

Cincinnati 11.18 

Chicago 11.07 

St.  Louis 10.55 

San  Francisco...     8.45 

New  Orleans 10.56 

Washington 11.48 

Charleston 11.36 

Havana 11.25 


AT   NEW  YORK. 

Boston 12.12  p.  M. 

Quebec 12.12 

Portland 12.15 

London 4.55 

Paris 5.05 

Rome 5.45 

Constantinople     6.41 

Vienna 600 

St.  Petersburg..     6.57 
Pekin,    night...  12.40A.M. 


TROT  WEIGHT. 

By  this  weight  gold,  silver,  platina  and  precious 
etones,  except  diamonds,  are  estimated. 

*20  Mites 1  Grain. 

20  Grains....  1  Penny wt. 

Any  quantity  of  gold  is  supposed  to  be  divided 


20  Punnywts 1  Ounce. 

12  Ounces 1  Pound. 


176        ORTON'S  LIGHTNING  CALCULATOR. 

into  24  parts,  called  carats.  If  pure,  it  is  said  to 
be  24  carats  fine;  if  there  be  22  parts  of  pure  gold 
and  2  parts  of  alloy,  it  is  said  to  be  22  carats  fine 
The  standard  of  American  coin  is  nine-tenths  pure 
gold,  and  is  worth  $20.67.  What  is  called  the 
new  standard,  used  for  watch  cases,  etc.,  is  18  carats 
fine.  The  term  carat  is  also  applied  to  a  weight  of 
3J  grains  troy,  used  in  weighing  diamonds ;  it  is 
divided  into  4  parts,  called  grains  ;  4  grains  troy 
are  thus  equal  to  5  grains  diamond  weight. 
APOTHECARIES'  WEIGHT — USED  IN  MEDICAL  PRESCRIPTIONS. 
The  pound  and  ounce  of  this  weight  are  the 
same  as  the  pound  and  ounce  troy,  but  differently 
divided. 
90  Grains  Troy...l  Scruple. 


8  Scruples 1  Drachm. 


8  Drachms...!  Ounce  Troy. 
12  Ounces....!  Pound  Troy. 


Druggists  buy  their  goods  by  avoirdupois  weight. 

AVOIRDUPOIS    WEIGHT. 

By  this  weight  all  goods  are  sold  except  those 
named  under  troy  weight. 


Grains  ................................  1  Dram. 

16'    Drama  .................................  1  Ounce. 

16     Ounces  .................................  1  Pound. 

28  Pounds  ...................................  1  Quarter. 

4  Quarters  or  100  pounds  ............  1  Hundred   Weight. 

20  Hundredweight  ......................  1  Ton. 

The  grain  avoirdupois,  though  never  used,  is  the 
same  as  the  grain  in  troy  weight.  7,000  grains 
make  the  avoirdupois  pound,  and  5,760  grains  tha 


WEIGHTS  AND  MEASURES.  177 

troy  pound.  Therefore,  the  troy  pound  is  less  than 
the  avoirdupois  pound  in  the  proportion  of  14  to 
17,  nearly;  but  the  troy  ounce  is  greater  than  the 
avoirdupois  ounce  in  the  proportion  of  79  to  72, 
nearly.  In  times  past  it  was  the  custom  to  allow 
112  pounds  for  a  hundred  weight,  but  usage,  as 
well  as  the  laws  of  a  majority  of  the  States,  at  the 
present  time  call  100  pounds  a  hundred  weight. 

APOTHECARIES'  FLUID  MEASURE. 

60  Minims 1  Fluid  Drachm. 

8  Fluid  Drachms 1  Ounce  (Troy). 

16  Ounces  (Troy) 1  Pint. 

8  Pints 1  Gallon. 

MEASURE  OP  CAPACITY  FOR  ALL  LIQUIDS. 

6  Ounces  Avoirdupois  of  water  make  1  Gill. 

4  Gills IPint       =  34f  Cubic  Inches  (nearly). 

2  Pints 1  Quart    =  69J  do 

4  Quarts 1  Gallon  =277J          do 

31J  Gallons 1  Barrel, 

42  Gallons 1  Tierce. 

63  Gallons,  or  2  bbls 1  Hogshead. 

2  Hogsheads 1  Pipe  or  Butt, 

2  Pipes 1  Tun. 

The  gallon  must  contain  exactly  10  pounds  avoir- 
dupois, of  pure  water,  at  a  temperature  of  62  °, 
the  barometer  being  at  30  inches.  It  is  the 
standard  unit  of  measure  of  capacity  for  liquids 
and  dry  goods  of  every  description,  and  is  J  larger 
than  the  old  wine  measure,  -gT2  larger  than  the  old 


178  ORTON  3  LIGHTNING  CALCULATOE. 

dry  measure,  and  ^  less  than  the  old  ale  measure 
The  wine  gallon  must  contain  231  cubic  inches. 

MEASURE  OF   CAPACITY   FOR  ALL  DRY  GOODS. 

4  Gills 1  pint        =     34f  cubic  inchs( nearly) 

2  Pints 1  quart      =     69J  cubic  inches. 

4  Quarts 1  gallon     =  27 7 J  cubic  inches. 

2  Gallons 1  peck       =  554J  cubic  inches. 

4  Pecks,  or  8  gals.  1  bushel    =2150 J  cubic  inches. 

8  Bushels 1  quarter  =     10 J  cubic  feet  (nearly). 

When  selling  the  following  articles  a  barrel 
weighs  as  here  stated : 

For  rice,  600  Ibs.;  flour,  196  Ibs.;  powder,  25 
Ibs.;  corn,  as  bought  and  sold  in  Kentucky,  Ten- 
nessee, etc.,  5  bushels  of  shelled  corn — as  bought 
and  sold  at  New  Orleans,  a  flour-barrel  full  of  ears: 
potatoes,  as  sold  in  New  York,  a  barrel  contains  2| 
bushels;  pork,  a  barrel  is  200  Ibs.,  distinguished 
in  quality  by  "clear,"  "mess,"  "prime;"  a  barrel 
of  beef  is  the  same  weight. 

The  legal  bushel  of  America  is  the  old  Win- 
chester measure  of  2,150.42  cubic  inches.  The 
imperial  bushel  of  England  is  2,218.142  cubic 
inches,  so  that  32  English  bushels  are  about  equa- 
to  33  of  ours. 

Although  we  are  all  the  time  talking  about  the 
price  of  grain,  etc.,  by  the  bushel,  we  sell  by 
weight,  as  follows : 

Wheat,  beans,  potatoes,  and  clover-seed,  60  Ibs 


WEIGHTS  AND  MEASURES.  179 

to  the  bushel ;  corn,  rye,  flax-seed,  and  onions,  56 
Ibs.;  corn  on  the  cob,  70  Ibs.;  buckwheat,  52  Ibs.; 
barley  48  Ibs.;  hemp-seed,  44  Ibs.j  timothy-seed, 
45  Ibs.;  castor  beans,  46  Ibs.;  oats,  35  Ibs.;  bran, 
20  Ibs.;  blue-grass  seed,  14  Ibs.;  salt — the  real 
weight  of  coarse  salt  is  85  Ibs.;  dried  apples,  24 
Ibs.;  dried  peaches,  33  Ibs.,  according  to  some 
rules,  but  others  are  22  Ibs.  for  a  bushel,  while 
tn  Indiana,  dried  apples  and  peaches  are  sold  by 
the  heaping  bushel ;  so  are  potatoes,  turnips, 
onions,  apples,  etc.,  and  in  some  sections  oats  are 
heaped.  A  bushel  of  corn  in  the  ear  is  three  heaped 
half  bushels,  or  four  even  full. 

In  Tennessee  a  hundred  ears  of  corn  is  some- 
times counted  as  a  bushel. 

A  hoop  18J  inches  diameter,  8  inches  deep, 
holds  a  Winchester  bushel.  A  box,  12  inches 
eauare,  7  and  7^  deep,  will  hold  half  a  busl  el 
A  heaping  bushel  is  2,815  cubic  inches. 

CLOTH  MEASURE. 

2J-  Inches 1  nail. 

4  Nails 1  quarter  of  a  yard, 

4  Quarters 1  yard. 

FOREIGN   CLOTH   MEASURE. 

2|  Quarters 1  Ell  Hamburgh. 

3  Quarters 1  Ell  Flemish. 

5  Quarters 1  Ell  English 

6  Quarters 1  Ell  French, 

12 


180        ORTON'S  LIGHTNING  CALCULATOR. 

MEASURE  OF  LENGTH. 

12  Inches  ....................  1  foot. 

3  Feet  ......................  1  yard. 

5J  Yards  ...................  1  rod,  pole,  or  perch, 

40  Poles  .....................  1  furlong. 

8  Furlongs,  or  1,760  yds,  1  mile. 


69-1-  Miles  )    1  degree  of  a  great  cir< 

'*  j 


--        es 

'*  ele  of  the  earth. 


By  scientific  persons  and  revenue  officers,  the 
Inch  is  divided  into  tenths,  liundredths^  etc.  Among 
mechanics,  the  inch  is  divided  into  eighths.  The 
division  of  the  inch  into  12  parts,  called  lines,  is 
not  now  in  use. 

A  standard  English  mile,  which  is  the  measure 
that  we  use,  is  5,280  feet  in  length,  1,760  yards,  or 
320  rods.  A  strip,  one  rod  wide  and  one  mile 
long,  is  two  acres.  By  this  it  is  easy  to  calculate 
the  quantity  of  land  taken  up  by  roads,  and  also 
how  much  is  wasted  by  fences. 

GUNTER'S  CHAIN. 

USED  FOR  LAND  MEASURE 

7  /^Inches  ..............................   1  Link. 

100  Links,  or  66  feet,  or  4  poles  ........  1  Chain. 

10  Chains  long  by  1  broad,  or  10  )     .,     A 

*?    -J  >    1   Acre. 

square  chains  ...................  j 

80  Chains  ................  ...................  1  Mile. 


WEIGHTS  AND  MEASURES.  181 

SURFACE   MEASURE. 

144  Sq.  inches  1  sq.  foot          I    40  Sq.  perches  1  rood 

9  Sq.  feet     1  sq.  yard  |      4  Roods 1  acre 

80^  Sq.  yards  1  sq.  rdorprch  j  640  Acres  1  sq.  mile 

Measure  209  feet  on  each  side,  and  you  have  a 
square  acre,  within  an  inch. 

The  following  gives  the  comparative  size,  in 
square  yards,  of  acres  in  different  countries  : 

English  acre,  4,840  square  yards ;  Scotch,  6,150; 
Irish,  7,840;  Hamburgh,  11,545;  Amsterdam, 
9,722;  Dantzic,  6,650 ;  France  (hectare),  11,960 ; 
Prussia  (morgen),  3,053. 

This  difference  should  he  borne  in  mind  in  read- 
ing of  the  products  per  acre  in  different  countries. 
Our  land  measure  is  that  of  England. 

GOVERNMENT  LAND  MEASURE. 

A  Township — 36  sections,  each  a  mile  square. 

A  section — 640  acres. 

A  quarter  section,  half  a  mile  square — 160  acres. 

An  eighth  section,  half  a  mile  long,  north  and 
south,  and  a  quarter  of  a  mile  wide — 80  acres. 

A  sixteenth  section,  a  quarter  of  a  mile  square — 
40  acres. 


182        ORTON'S  LIGHTNING  CALCULATOR. 


The  sections  are  all    numbered  1  to  36,  com- 
mencing at  the  north-east  corner,  thus : 


6 

5 

4 

3 

2 

NWiX  E 
SWJSE 

7 

8 

9 

10 

11 

12 

18 

17 

16* 

15 

14 

13 

19 

20 

21 

22 

23 

24 

30 

29 

28 

27 

26 

25 

31 

32 

33 

34 

35 

36 

The  sections  are  all  divided  in  quarters,  which 
are  named  by  the  cardinal  points,  as  in  section  1. 
The  quarters  are  divided  in  the  same  way.  The 
description  of  a  forty-acre  lot  would  read:  The 
south  half  of  the  west  half  of  the  south-west 
quarter  of  section  1  in  township  24,  north  of  range 
7  west,  or  as  the  case  might  be ;  and  sometimes 
will  fall  short,  and  sometimes  overrun  the  number 
of  acres  it  is  supposed  to  contain. 

SQUARE  MEASURE — FOR  CARPENTERS,  MASONS,  ETC 

144  Sq  Inches 1  Sq  Foot. 

9  Sq  Ft,  or  1,296  Sq  In.  1  Sq  Yard, 
100  Sq  Feet 1  Sq  of  Flooring,  Roofing,  eta 

30J  Sq  Yards 1  Sq  Rod. 

86  Sq  Yards 1  Rood  of  Building. 

School  section. 


WEIGHTS  AND  MEASURES. 
GEOGRAPHICAL  OR  NAUTICAL  MEASURE. 

6  Feet  ..................  1  Fathom. 

110  Fathoms  or  660  ft.  1  Furlong. 
6075|   Feet  ...............  1  Nautical  Mile. 

3  Nautical  Miles  ......  1  League. 


-n  f    The  earth's  circumference 

.Degrees  ..........  -j  o^  OKKI      -i 

(      =24,855  \  miles,  nearly. 

The  nautical  mile  is  795|  feet  longer  than  the 
common  mile. 

MEASURE  OF  SOLIDITY. 

1728  Cubic  Inches  .............  1  Cubic  Foot, 

27  Cubic  Feet  .................  1  Cubic  Yard. 

16  Cubic  Feet  .................  1  Cord  Foot,  or  a  ft  of  wood. 

8  Cord  ft  or  128  Cubic  ft..  1  Cord. 
40  ft  of  round  or  50  ft 
of  hewn  timber.. 

42   Cubic  Feet  ...............  1  Ton  of  Shipping. 


0  ft  \    n  T 
r...  /   J 


ANGULAR  MEASURE,  OR  DIVISIONS  OF  THE  CIRCLE. 


60  Seconds 1  Minute. 

60  Minutes 1  Degree. 


360  Degrees. 


>  Degrees 1  Sign. 

90  Degrees 1  Quadrant, 

1  Circumference. 

MEASURE  OF  TIME. 

60  Seconds 1  Minute 

60  Minutes 1  Hour. 

24  Hours 1  Day. 

7  Days 1  Week. 

28  Days , 1  Lunar  Month., 

28  29,  30  or  31  Days 1  Cal.  Month. 

12  Cal.  Months 1  Year. 

865  Days 1  Com.  Year. 

SG6  Days 1  Leap  Year. 

S65|-  Days 1  Julian  Year. 

365  D.,  5  H.,  48  M.,  49  s 1  Solar  Year. 

365  D.,  6  H.,    9  M.,  12  s 1  Siderial  Year 

Q 


I 

184       ORTON'S  LIGHTNING  CALCULATOR. 

ROPES  AND  CABLES. 

6  Feet 1  Fathom 

120  Feet 1  Cable  Length. 

Miscellaneous  Important  Facts  about  Weights  and 
Measures. 

BOARD  MEASURE. 

Boards  are  sold  by  superficial  measure,  at  so 
much  per  foot  of  one  inch  or  less  in  thickness, 
adding  one-fourth  to  the  price  for  each  quarter* 
inch  thickness  over  an  inch. 

GRAIN  MEASURE  IN  BULK. 

Multiply  the  width  and  length  of  the  pile  to- 
gether, and  that  product  by  the  hight,  and  divide 
by  2,150,  and  you  have  the  contents  in  bushels. 

If  you  wish  the  contents  of  a  pile  of  ears  of 
corn,  or  roots,  in  heaped  bushels,  ascertain  the 
cubic  inches  and  divide  by  2,818. 

A  TON  WEIGHT. 

In  this  country  a  ton  is  2,000  pounds.  In  most 
places  a  ton  of  hay,  etc.,  is  2,240  pounds,  and  in 
Borne  places  that  foolish  fashion  still  prevails  of 
weighing  all  bulky  articles  sold  by  the  tun,  by  the 
"  long  weight,"  or  tare  of  12  Ibs.  per  cwt. 

A  tun  of  round  timber  :"s  40  feet;  of  square 
timber,  54  cubic  feet. 


WEIQHTS  AND  MEASURES.  185 

A  quarter  of  corn  or  other  grain  sold  by  the 
bushel  Is  eight  imperial  bushels,  or  quarter  of  a 
tun. 

A  ton  of  liquid  measure  is  252  gallons. 

BUTTER 

Is  so\d  by  avoirdupois  weight,  which  compares  with 
troy  weight  as  144  to  175 ;  the  troy  pound  being 
that  much  the  lightest.  But  175  troy  ounces 
equal  192  of  avoirdupois. 

A  firkin  of  butter  is-  56  Ibs.;  a  tub  of  butter  is 
84  Ibs. 

THE    KILOGRAMME   OP   FRANCE 

Is  1000  grammes,  and  equal  to  2  Ibs.  2  oz.  4  grs. 
avoirdupois. 

A  BALE  OF  COTTON, 

In  Egypt,  is  90  Ibs.;  in  America  a  commercial  bale 
is  400  Ibs.;  though  put  up  to  vary  from  280  to  720, 
in  different  localities. 

A  bale  or  bag  of  Sea  Island  cotton  is  300  Ibs. 

WOOL. 

In  England,  wool  is  sold  by  the  sack  or  boll,  of 
22  stones,  which,  at  14  Ibs.  the  stone,  is  308  Ibs. 

A  pack  of  wool  is  17  stone,  2  Ibs.,  which  is  rated 
as  a  pack  load  for  a  horse.  It  is  240  Ibs.  A  tod 
of  wool  is  2  stones  of  14  Ibs.  A  wey  of  wool  is  6J 
tods.  Two  weys,  a  sack.  A  clove  of  wool  is  half 
&  stone. 


186       ORTON'S  LIGHTNING  CATCULATOB. 

THE  STONE  WEIGHT 

So  often  spoken  of  in  English  measures,  is  14 
Ibs.,  when  weighing  wool,  feathers,  hay,  etc.;  but 
a  stone  of  beef,  fish,  butter,  cheese,  etc.,  is  only  8 
pounds. 

HAY. 

In  England,  a  truss,  when  new,  is  60  Ibs.,  or  56 
Ibs.  of  old  hay.  A  truss  of  straw,  40  Ibs.  A  load 
of  hay  is  36  trusses. 

In  this  country,  a  load  is  just  what  it  may  hap- 
pen to  weigh ;  and  a  tun  of  hay  is  either  2,000 
Ibs.  or  2,240  Ibs.,  according  to  the  custom  of  the 
locality.  A  bale  of  hay  is  generally  considered 
about  300  Ibs.,  but  there  is  no  regularity  in  the 
weight.  A  cube  of  a  solid  mow,  10  feet  square, 
will  weigh  a  tun. 

A  LAST 

Is  an  English  measure  of  various  articles. 

A  last  of  soap,  ashes,  herrings,  and  some  snail 
lar  things,  is  2  barrels. 

A  last  of  corn  is  10  quarters. 

A  last  of  gunpowder,  24  barrels. 

A  last  of  flax  or  feathers,  1,700  Ibs. 

A  last  of  wool,  12  sacks. 

A  SCOTCH   PINT 

Contains  105  cubic  inches,  and  is  equal  to  foul 
English  pints.  21 J  Scotch  pints  make  a  farlot  of 
wheat. 


Tf EIGHTS  AND  MEASURES.  187 

COAL. 

A  chaldron  is  58f  cubic  feet,  or  by  measure,  36 
heaped  bushels.  A  heaped  bushel  of  anthracite 
coal  weighs  80  Ibs.,  making  2,880  Ibs.  to  a 
chaldron. 

WOOD. 

A  cord  of  wood  is  128  solid  feet,  in  this  country 
and  England.  In  France  it  is  576-feet.  We  cord 
wood  4  feet  long,  in  piles  4  feet  by  8. 

In  New  Orleans  wood  is  retailed  by  the  pound, 
and  to  a  limited  extent  here.  It  is  also  sold  by 
the  barrel.  A  load  of  wood  in  New  York  is  42| 
cubic  feet,  or  one-third  of  a  cord. 

Wood  is  sold  in  England  by  the  stack,  skid, 
quintal,  billet,  and  bundle. 

A  stack  is  108  solid  feet,  and  unusually  piled  12 
feet  long,  3  feet  high,  and  3  feet  wide. 

A  quintal  of  wood  is  100  Ibs. 

A  skid  is  a  round  bundle  of  sticks,  4  feet  long 
A  one-notch  skid  girts  16  inches.  A  two-notch 
skid,  23  inches.  A  three-notch  skid,  28  inches. 
A  four-notch  skid,  33  inches.  A  five-notch  skid, 
38  inches. 

A  billet  of  wood  is  a  bundle  of  sticks,  3  feet 
long,  and  girts  7,  10,  or  14  inches,  and  these  bun- 
dles sell  by  the  score  or  hundred.  A  score  is  20, 
and  comes  from  the  count  by  tally,  or  marks. 


188 


ORTON'S  LIGHTNING  CALCULATOR. 


Faggots  of  wood  are  bundles  of  brush  3  feet 
long,  two  round.     A  load  of  faggots  is  50  bundles, 
All  wood  should  be  sold  by  the  pound. 

CAPACITY  OF  CISTERNS  OR  WELLS. 

Tabular  view  of  the  number  of  gallons  contained 
in  the  clear  between  the  brickwork  for  each  ten 
inches  of  depth : 


DT4.ME' 

2  f  e 

f 

9 

9. 
9 

?. 

71  < 

PER. 
et  eqi 

i 

GALLONS. 

ial  19 

DIAME 

8  f  e 
98} 
!90} 

11 

12 
13 
14 

15     < 
20     < 
25     ' 

TER. 

et  eqi 

i 

i 
i 
i         < 

GALLONS. 

ial  313 

30 

353 

44 

396 

60 

461 

78 

489 

97 

692 

122 

705 

148 

827 

176 

959 

207 

1101 

240 

1958 

..  275 

<    .         ..  3059 

TO  MEASURE  CORN  IN  THE  CRIB. 

Corn  is  generally  put  up  in  cribs  made  of  rails, 
but  the  rule  will  apply  to  a  crib  of  any  size  or 
kind. 

Two  cubic  feet  of  good,  sound,  dry  corn  in  th* 
ear,  will  make  a  bushel  of  shelled  corn.  To  get 
then,  the  quantity  of  shelled  corn  in  a  crib  of  core 
in  the  ear,  measure  the  length,  breadth,  and  high* 
of  the  crib,  inside  the  rail;  multiply  the  length  by 


WEIGHTS  AND  MEASURES.  189 

the  breadth,  and  the  product  by  the  night ;  then 
divide  the  result  by  two,  and  you  have  the  number 
of  bushels  of  shelled  corn  in  the  crib. 

In  measuring  the  hight,  of  course,  the  hight  of 
the  corn  is  intended.  And  there  will  be  found  to 
be  a  difference  in  measuring  corn  in  this  mode, 
between  fall  and  spring,  because  it  shrinks  very 
much  in  the  winter  and  spring,  and  settles  down. 

KULES   FOR   DETERMINING   THE    WEIGHT    OF    LIVE 
CATTLE. 

Measure  in  inches  the  girth  round  the  breast, 
just  behind  the  shoulder-blade,  and  the  length  of 
the  back  from  the  tail  to  the  forepart  of  the  shoul- 
der-blade. Multiply  the  girth  by  the  length,  and 
divide  by  144.  If  the  girth  is  less  than  three  feet, 
multiply  the  quotient  by  11 ;  if  between  three  feet 
and  five  feet,  multiply  by  16 ;  if  between  five  feet 
and  seven  feet,  multiply  by  23  ;  if  between  seven 
and  nine  feet,  multiply  by  31.  If  the  animal  is 
lean,  deduct  ^  from  the  result. 

Take  the  girth  and  length  in  feet,  multiply  the 
square  of  the  girth  by  the  length,  and  multiply 
the  product  by  3.36.  The  result  will  be  the  an- 
swer in  pounds.  The  live  weight,  multiplied  by 
.605  gives  a  near  approximation  to  the  net  weight 


190       ORTON'S  LIGHTNING  CALCULATOR. 


ASTRONOMICAL  CALCULATIONS. 

A  scientific  method  of  telling  immediately  what  day 
of  the  wecJc  any  date  transpired  or  will  transpirej 
from  the  commencement  of  the  Christian  Era*  for 
the  term  of  three  thousand  years. 

MONTHLY   TABLE. 

The  ratio  to  add  for  each  month  will  be  found 
in  the  following  table: 


Ratio  of  June  is 0 

Ratio  of  September  is 1 

Ratio  of  December  is.......l 

Ratio  of  April  is 2 

Ratio  of  July  is 2 

Ratio  of  January  is 3 


Ratio  of  October  is 3 

Ratio  of  May  is 4 

Ratio  of  August  is 5 

Ratio  of  March  is 6 

Ratio  of  February  is 6 

Ratio  of  November  is 6 


NOTE. — On  Leap  Year  the  ratio  of  January  is  2,  and 
the  ratio  of  February  is  5.  The  ratio  of  the  other  ten 
months  do  not  change  on  Leap  Years. 

CENTENNIAL   TABLE. 

The  ratio  to  add  for.  each  century  will  be  found 
in  the  following  table: 

Q    200,     900,  1800,  2200,  2600,  3000,  ratio  is 0 

§!    300,  1000,  ratio  is 6 

|    400,  1100,  1900,  2300,  2700,  ratio  is 5 

*     600   1200,  1600,  2000,  2400,  2800,  ratio  is 4 

3     600    1300,  ratio  is 3 

000,  700,  1400,  1700,  2100,  2500,  2900,  ratio  is 2 

100,  800,  1500,  ratio  is 1 


ASTRONOMICAL  CALCULATIONS.      191 

NOTE. — The  figure  opposite  each  century  is  its  ratio; 
thus  the  ratio  for  200,  900,  etc.,  is  0.  To  find  the  ratio 
of  any  century,  first  find  the  century  in  the  above  table, 
then  run  the  eye  along  the  line  until  you  arrive  at  the 
end;  the  small  figure  at  the  end  is  its  ratio. 

METHOD   OP   OPERATION. 

RULE.* — To  the  given  year  add  its  fourth  part} 
rejecting  the  fractions  ;  to  this  sum  add  the  day  of 
the  month;  then  add  the  ratio  of  the  month  and 
the  ratio  of  the  century.  Divide  this  sum  by  7 ;  the 
remainder  is  the  day  of  the  week,  counting  Sunday 
as  the  first,  Monday  as  the  second,  Tuesday  as  the 
third,  Wednesday  as  the  fourth,  Thursday  as  the 
fifth,  Friday  as  the  sixth,  Saturday  as  the  seventh; 
the  remainder  for  Saturday  will  be  0  or  zero. 

EXAMPLE  1. — Required  the  day  of  the  week 
for  the  4th  of  July,  1810. 

To  the  given  year,  which  is 10 

Add  its  fourth  part,  rejecting  fractions 2 

Now  add  the  day  of  the  month,  which  is 4 

Now  add  the  ratio  of  July,  which  is...  2 

Now  add  the  ratio  of  1800,  which  is..( 0 

Divide  the  .whole  sum  by  7.  7  |  J.8 — 4 

~2 

We  have  4  for  a  remainder,  which  signifies  the 
fourth  day  of  the  week,  or  Wednesday. 

*When  dividing  the  year  by  4,  always  leave  off  the  centuries.  Va 
divide  by  4  to  find  the  number  of  Leap  Years. 


192       ORTON'S  LIGHTNING  CALCULATOR. 

NOTE. — In  finding  the  day  of  the  week  for  the  present 
century,  no  attention  need  be  paid  vo  the  centennial  ratio^ 
as  it  is  0. 

V 

EXAMPLE  2. — Required  the  day  of  the  week 
for  the  2d  of  June,  1805. 

To  the  given  year,  which  is 5 

Add  its  fourth  part,  rejecting  fractions 1 

Now  add  the  day  of  the  month,  which  is 2 

Now  add  the  ratio  of  June,  which  is 0 

Divide  the  whole  sum  by  7.  7  |  8—1 

i 

"We  have  1  for  a  remainder,  which  signifies  the 
first  day  of  the  week,  or  Sunday. 

The  Declaration  of  American  Independence 
was  signed  July  4,  1776.  Required  the  day  of 
the  week. 

To  the  given  year,  which  is 76 

Add  its  fourth  part,  rejecting  fractions 19 

Now  add  the  day  of  the  month,  which  is 4 

Now  add  the  ratio  of  July,  which  is 2 

Now  add  the  ratio  of  1700,  which  is 2 

Divide  the  whole  sum  by  7.  7  |  103—5 

~14 

TVe  have  5  for  a  remainder,  which  signifies  the 
fifth  day  of  the  week,  or  Thursday. 

The  Pilgrim  Fathers  landed  on  Plymouth  Rock 
Dec.  20,  1620.  Required  the  day  of  the  week. 


ASTRONOMICAL   CALCULATIONS.  193  . 

To  the  given  year,  which  is 20 

Add  its  fourth  part,  rejecting  fractions 5 

Now  add  the  day  of  the  month,  which  is 20 

Now  add  the  ratio  of  December,  which  is 1 

Now  add  the  ratio  of  1600,  which  is 4 

Divide  the  whole  sum  by  7.  7  |_50— 1   * 

7 

We  have  1  for  a  remainder,  which  signifies  tha 
first  day  of  the  week,  or  Sunday. 

On  what  day  will  happen  the  8th  of  January, 
1815?  An*.  Sunday. 

On  what  day  will  happen  the  4th  of  May,  1810? 

On  what  day  will  happen  the  3d  of  December, 
1423?  An*.  Friday. 

On  what  day  of  the  week  were  you  born? 

The  earth  revolves  round  the  sun  once  in  365 
days,  5  hours,  48  minutes,  48  seconds;  this  period 
is,  therefore,  a  Solar  year.  In  order  to  keep  pace 
with  the  solar  year,  in  our  reckoning,  we  make 
every  fourth  to  contain  366  days,  and  call  it  Leap 
Year.  Still  greater  accuracy  requires,  however, 
that  the  leap  day  be  dispensed  with  three  times 
in  every  400  years.  Hence,  every  year  (except 
the  centennial  years)  that  is  divisible  by  4  is  a 
Leap  Year,  and  every  centennial  year  that  is 
divisible  by  400  is  also  a  Leap  Year.  The  next 
centennial  year  that  will  be  a  Leap  Year  h  2000 


194          ORTON'S   LIGHTNING  CALCULATOR. 

For  the  practical  convenience  of  those  who  have  occasion  to 

refer  to  mensuration,  we  have  arranged  the  following  useful  .able 

of  multiples.    It  covers  the  whole  ground  of  practical  geometry, 

and  should  be  studied  carefully  by  those  who  wish  to  be  skilled  in 

this  beautiful  branch  of  mathematics: 

TABLE  OF  MULTIPLES.' 

Diameter  of  a  circle  X  3-1416  —  Circumference. 

Radius  of  a  circle  X  6.283185  —  Circumference. 

Square  of  the  radius  of  a  circle  X  3.1416  —  Area. 

Square  of  the  diameter  of  a  circle  X  0.7854  —  Area. 

Bquare  of  the  circumference  of  a  circle  X  0.07958  =  Area. 

Half  the  circumference  of  a  circle  X  D7  nalf  its  diameter  —  Area, 

Circumference  of  a  circle  X  0.159155  —  Radius. 

Bquare  root  of  the  area  of  a  circle  X  0.56419  —  Radius. 

Circumference  of  a  circle  X  0.31831  —  Diameter. 

Bquare  root  of  the  area  of  a  circle  X  1.12838  —  Diameter. 

Diameter  of  a  circle  X  °.86  —  Side  of  inscribed  equilateral  triangle. 

Diameter  of  a  circle  X  0.7071  —  Side  of  an  inscribed  square. 

Circumference  of  a  circle  X  °-225  ™  Sid®  of  *&  inscribed  square. 

Circumference  of  a  circle  X  0.282  —  Side  of  an  equal  square. 

Diameter  of  a  circle  X  0.8862  —  Side  of  an  equal  square. 

Base  of  a  triangle  X  by  %  *ne  altitude  —  Area. 

Multiplying  both  diameters  and  .7854  together  —  Area  of  an  ellipse. 

Surface  of  a  sphere  X  bv  %  of  its  diameter  —  Solidity. 

Circumference  of  a  sphere  X  D7  ^s  diameter  —  Surface. 

Square  of  the  diameter  of  a  sphere  X  3.1416  -=  Surface. 

Square  of  the  circumference  of  a  sphere  X  0.3183  =  Surface. 

Cube  of  the  diameter  of  a  sphere  X  0.5236  —  Solidity. 

Cube  of  the  radius  of  a  sphere  X  4.1888  —  Solidity. 

Cube  of  the  circumference  of  a  sphere  X  0.016887  —  Solidity. 

Square  root  of  the  surface  of  a  sphere  X  0.56419  —  Diameter. 

Square  root  of  the  surface  of  a  sphere  X  1.772454  —  Circumference 

Cube  root  of  the  solidity  of  a  sphere  X  1-2407  —  Diameter. 

Cube  root  of  the  solidity  of  a  sphere  X  3.8978  —  Circumference, 

Radius  of  a  sphere  X  1.1547  -=»  Side  of  inscribed  cube. 

Bquare  root  of  (%  of  the  square  of)  the  diameter  of  a  sphere  -• 

Side  of  inscribed  cube, 
irea  of  its  tase  X  by  %  of  its  altitude  —  Solidity  of  a  cone  or  pyr- 

amid,  whether  round,  square,  or  triangular. 
4rea  of  one  of  its  sides  X  6  =  Surface  of  a  cube. 
41Utu.de  of  Tiapezoid  X  V*  the  sum  of  its  parallel  aides  —  Area. 


i 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  PINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


YA  02396 


M306O26 

Q 

07 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


CONTENTS. 

1.  THREE  NEW  PROCESSES  OF  ADDITION. 

2.  Two  PROCESSES  OF  PROVING  SUBTRACTION. 

3.  LIGHTKINC  METHOD  OF  MULTIPLICATION. 

4.  RAP.D  PROCESS  OF  SQUARING  AND  CUBING. 

5.  THREE  NEW  FORMS  o*  CONTRACTIONS  IN  DIVISION. 

6.  MENTAL  OPERATIONS  -x  FRACTJ. 

7.  THE  CANCELING  SYSTEM  THOROUGHLY  EXPLAINED. 

8.  LIGHTNING  METHOD  OF  COMPUTING  INTEREST. 

9.  PARTIAL  PAYMENTS  ON  NOTES  AND^BONDS. 
Vwo  PROCESSES  OF  AVERAGING  ACCOUNTS. 

11.  EXTRACTIONS  OF  THE  ROOTS  MENTALLY. 

12.  To  MEASURE  CORN  IN  THE  CRIB. 

13.  "^\pir»  PROCESS  OF  MEASURING  WOOD,  BARK,  OR  COAL. 

14.  ..-.PII*  PROCESS  OF  MEASURING  ALL  KINDS  OF  TIMBER. 

15.  HOW  TO  TELL  THE  No.   OP  INCH  BOARDS  IN  A  LOG. 

16.  WEIGHTS  AND  MEASURES. 

17.  ASTRONOMICAL  CALCULATE 

MULTIPLICATION 

- 

:  D  VALUABLE  INFORMATI-X   TH  BOOH-K 


peculiarities  which  tl.s  student  will  here  fin'd  in 

,    Interest,    Mensuration,    and    in    Averagin 

-.',  not  all  new.     Indeed  there  can  be  nothini 

;-le  ;  but,  as  far  as  the  author's  knowledge  ex 

i ware  that  these  A!  r  bet 

.   any  Arithmetical  work.     The  >n  se« 

uthat  the  poople  could  not  corn 

preciate  Mathematical  beauty 
otherwise,  presents   this   brie' 

.  with  the  full  assurance  that  w' 
pay  vine  attentior  to  the  subject  will  l>-.; 
abundantly  i> 


